Chapter 11 of RD Sharma's Class 11 Mathematics textbook delves into the fascinating world of trigonometric equations. These equations involve trigonometric functions and are fundamental in various fields such as physics, engineering, and mathematics. In Exercise 11.1 students are introduced to solving basic trigonometric equations building a strong foundation for the more complex problems in the later exercises.
Trigonometric Equations
Trigonometric equations are mathematical expressions that equate two trigonometric functions often involving angles. The solutions to these equations are the angles that satisfy the given equation. The Common trigonometric equations include those involving sine, cosine, tangent, and their reciprocal functions. Solving these equations typically requires knowledge of the trigonometric identities and properties, as well as a solid understanding of the unit circle.
Question 1. Find the general solution of the following equations:
(i) sin θ = 1/2
Solution:
We are given,
=> sin θ = 1/2
=> sin θ = sin π/6
We know if sin θ = sin a, the general solution is given by, θ = nπ + (−1)n a; n ∈ Z.
=> θ = nπ + (−1)n (π/6) ; n ∈ Z
(ii) cos θ = −√3/2
Solution:
We are given,
=> cos θ = −√3/2
=> cos θ = cos (π + π/6)
=> cos θ = cos (7π/6)
We know if cos θ = cos a, the general solution is given by, θ = 2nπ ± a ; n ∈ Z.
=> θ = 2nπ ± (7π/6) ; n ∈ Z
(iii) cosec θ = −√2
Solution:
We are given,
=> cosec θ = −√2
=> 1/sin θ = −√2
=> sin θ = −1/√2
=> sin θ = −sin (π/4)
As sin (−θ) = − sin θ, we have,
=> sin θ = sin (−π/4)
We know if sin θ = sin a, the general solution is given by, θ = nπ + (−1)n a ; n ∈ Z.
=> θ = nπ + (−1)n+1 (π/4) ; n ∈ Z
(iv) sec θ = √2
Solution:
We are given,
=> sec θ = √2
=> 1/sec θ = √2
=> cos θ = 1/√2
=> cos θ = cos π/4
We know if cos θ = cos a, the general solution is given by, θ = 2nπ ± a ; n ∈ Z.
=> θ = 2nπ ± (π/4) ; n ∈ Z
(v) tan θ = −1/√3
Solution:
We are given,
=> tan θ = −1/√3
=> tan θ = −tan (π/6)
As tan (−θ) = − tan θ, we have,
=> tan θ = tan (−π/6)
We know if tan θ = tan a, the general solution is given by, θ = nπ + a ; n ∈ Z.
=> θ = nπ − (π/6) ; n ∈ Z
(vi) √3 sec θ = 2
Solution:
We are given,
=> √3 sec θ = 2
=> √3 (1/cos θ) = 2
=> cos θ = √3/2
=> cos θ = cos (π/6)
We know if cos θ = cos a, the general solution is given by, θ = 2nπ ± a; n ∈ Z.
=> θ = 2nπ ± (π/6) ; n ∈ Z
Question 2. Find the general solution of following equations:
(i) sin 2θ = √3/2
Solution:
We are given,
=> sin 2θ = √3/2
=> sin 2θ = sin (π/3)
We know if sin θ = sin a, the general solution is given by, θ = nπ + (−1)n a; n ∈ Z.
=> 2θ = nπ + (−1)n (π/3)
=> θ = nπ/2 + (−1)n (π/6) ; n ∈ Z
(ii) cos 3θ = 1/2
Solution:
We are given,
=> cos 3θ = 1/2
=> cos 3θ = cos (π/3)
We know if cos θ = cos a, the general solution is given by, θ = 2nπ ± a; n ∈ Z.
=> 3θ = 2nπ ± (π/3)
=> θ = 2nπ/3 ± (π/9) ; n ∈ Z
(iii) sin 9θ = sin θ
Solution:
We are given,
=> sin 9θ = sin θ
=> sin 9θ – sin θ = 0
=> 2 cos 5θ sin 4θ = 0
=> cos 5θ = 0 or sin 4θ = 0
We know, if cos θ = 0, then θ = (2n+1)π/2 and if sin θ = 0, then θ = nπ where n ∈ Z.
=> 5θ = (2n+1)π/2 or 4θ = nπ
=> θ = (2n+1)π/10 or θ = nπ/4, n ∈ Z
(iv) sin 2θ = cos 3θ
Solution:
We are given,
=> sin 2θ = cos 3θ
=> cos 3θ = sin 2θ
=> cos 3θ = cos (π/2 − 2θ)
We know if cos θ = cos a, the general solution is given by, θ = 2nπ ± a ; n ∈ Z.
=> 3θ = 2nπ ± (π/2 − 2θ)
=> 3θ = 2nπ + π/2 − 2θ or 3θ = 2nπ − π/2 + 2θ
=> 5θ = 2nπ + π/2 or θ = 2nπ − π/2
=> 5θ = (4n+1) (π/2) or θ = (4n−1) (π/2)
=> θ = (4n+1)π/10 or θ = (4n−1)π/2, n ∈ Z
(v) tan θ + cot 2θ = 0
Solution:
We are given,
=> tan θ + cot 2θ = 0
=> cot 2θ = − tan θ
=> tan 2θ = − cot θ
=> tan 2θ = − tan (π/2 − θ)
As tan (−θ) = − tan θ, we have,
=> tan 2θ = tan (θ − π/2)
We know if tan θ = tan a, the general solution is given by, θ = nπ + a ; n ∈ Z.
=> 2θ = nπ + θ − π/2
=> θ = nπ − π/2 ; n ∈ Z
(vi) tan 3θ = cot θ
Solution:
We are given,
=> tan 3θ = cot θ
=> tan 3θ = tan (π/2 − θ)
We know if tan θ = tan a, the general solution is given by, θ = nπ + a ; n ∈ Z.
=> 3θ = nπ + π/2 − θ
=> 4θ = nπ + π/2
=> θ = nπ/4 + π/8 ; n ∈ Z
(vii) tan 2θ tan θ = 1
Solution:
We are given,
=> tan 2θ tan θ = 1
=> tan 2θ = cot θ
=> tan 2θ = tan (π/2 − θ)
We know if tan θ = tan a, the general solution is given by, θ = nπ + a ; n ∈ Z.
=> 2θ = nπ + π/2 − θ
=> 3θ = nπ + π/2
=> θ = nπ/3 + π/6 ; n ∈ Z
(viii) tan mθ + cot nθ = 0
Solution:
We are given,
=> tan mθ + cot nθ = 0
=> sin mθ/cos mθ + cos nθ/sin nθ = 0
=> sin mθ sin nθ + cos nθ cos mθ = 0
=> cos (m−n)θ = 0
We know, if cos θ = 0, then θ = (2k+1)π/2 where k ∈ Z.
=> (m−n)θ = (2k+1)π/2
=> θ = (2k+1)π/2(m−n) ; n ∈ Z
(ix) tan pθ = cot qθ
Solution:
We are given,
=> tan pθ = cot qθ
=> tan pθ = tan (π/2 − qθ)
We know if tan θ = tan a, the general solution is given by, θ = nπ + a ; n ∈ Z.
=> pθ = nπ + π/2 − qθ
=> (p + q)θ = (2n+1)π/2
=> θ = (2n+1)π/2(p + q) ; n ∈ Z
(x) sin 2x + cos x = 0
Solution:
We are given,
=> sin 2x + cos x = 0
=> cosx = − sin 2x
=> cos x = cos (π/2 + 2x)
We know if cos θ = cos a, the general solution is given by, θ = 2nπ ± a ; n ∈ Z.
=> x = 2nπ ± (π/2 + 2x)
=> x = 2nπ + π/2 + 2x or x = 2nπ − π/2 − 2x
=> x = −(4n+1)π/2 or 3x = (4n−1)π/2
=> x = −(4n+1)π/2 or 3x = (4n−1)π/2
=> x = (4m−1)π/2, where m = −n or x = (4n−1)π/6, n ∈ Z
(xi) sin θ = tan θ
Solution:
We are given,
=> sin θ = tan θ
=> sin θ = sin θ/cos θ
=> sin θ (cos θ − 1) = 0
=> sin θ = 0 or cos θ = 1
=> sin θ = 0 or cos θ = cos 0
We know, if cos θ = 0, then θ = (2n+1)π/2 and if sin θ = 0, then θ = nπ where n ∈ Z.
=> θ = nπ or θ = 2nπ, n ∈ Z
(xii) sin 3x + cos 2x = 0
Solution:
We are given,
=> sin 3x + cos 2x = 0
=> cos 2x = −sin 3x
=>cos 2x = cos (π/2 + 3x)
We know if cos θ = cos a, the general solution is given by, θ = 2nπ ± a ; n ∈ Z.
=> 2x = 2nπ ± (π/2 + 3x)
=> 2x = 2nπ + π/2 + 3x or 2x = 2nπ − π/2 − 3x
=> −x = 2nπ + π/2 or 5x = 2nπ − π/2
=> x = 2mπ − π/2, where m = −n or x = (2nπ − π/2)/5
=> x = (4m−1)π/2, where m = −n or x = (4n−1)π/10, n ∈ Z
Question 3. Solve the following equations:
(i) sin2 θ − cos θ = 1/4
Solution:
We have,
=> sin2 θ − cos θ = 1/4
=> 1 − cos2 θ − cos θ = 1/4
=> cos2 θ + cos θ − 3/4 = 0
=> 4cos2 θ + 4cos θ − 3 = 0
=> 4cos2 θ + 6cos θ − 2cos θ − 3 = 0
=> 2cos θ (2cos + 3) − (2cos θ + 3) = 0
=> (2cos θ − 1) (2cos θ + 3) = 0
=> cos θ = 1/2 or cos θ = −3/2
Ignoring cos θ = −3/2 as −1 ≤ cosθ ≤ 1. So, we have, cos θ = 1/2.
=> cos θ = cos π/3
=> θ = 2nπ ± π/3 ; n ∈ Z
(ii) 2cos2 θ − 5cos θ + 2 = 0
Solution:
We are given,
=> 2cos2 θ − 5cos θ + 2 = 0
=> 2cos2 θ − 4cos θ − cos θ + 2 = 0
=> 2cos θ (cos θ − 2) − (cos θ − 2) = 0
=> (2cos θ − 1) (cos θ − 2) = 0
=> cos θ = 1/2 or cos θ = 2
Ignoring cos θ = 2 as −1 ≤ cosθ ≤ 1. So, we have, cos θ = 1/2.
=> cos θ = cos π/3
=> θ = 2nπ ± π/3 ; n ∈ Z
(iii) 2sin2 x + √3cos x + 1 = 0
Solution:
We are given,
=> 2sin2 x + √3cos x + 1 = 0
=> 2 (1 − cos2 x) + √3cos x + 1 = 0
=> 2cos2 x − √3cos x − 3 = 0
=> 2cos2 x − 2√3cos x + √3cos x − 3 = 0
=> 2cosx (cos x − √3) + √3(cos x − √3) = 0
=> (2cosx + √3) (cos x − √3) = 0
=> x = −√3/2 or x = √3
Ignoring cos x = √3 as −1 ≤ cosθ ≤ 1. So, we have, cos x = −√3/2.
=> cos x = − cos π/6
=> cos x = cos (π − π/6)
=> cos x = cos (5π/6)
=> x = 2nπ ± 5π/6 ; n ∈ Z
(iv) 4sin2 θ − 8cos θ + 1 = 0
Solution:
We are given,
=> 4sin2 θ − 8cos θ + 1 = 0
=> 4 (1 − cos2 θ) − 8cos θ + 1 = 0
=> 4 cos2 θ + 8 cos θ − 5 = 0
=> 4 cos2 θ + 10 cos θ − 2 cos θ − 5 = 0
=> 2cos θ (2cos θ +5) − (2cos θ +5) = 0
=> (2cos θ − 1) (2cos θ +5) = 0
=> cos θ = 1/2 or cos θ = −5/2
Ignoring cos x = −5/2 as −1 ≤ cosθ ≤ 1. So, we have, cos x = 1/2.
=> cos x = cos π/3
=> x = 2nπ ± π/3 ; n ∈ Z
(v) tan2 x + (1 − √3)tan x − √3 = 0
Solution:
We are given,
=> tan2 x + (1 − √3)tan x − √3 = 0
=> tan2 x + tan x − √3tan x − √3 = 0
=> tan x (tan x + 1) − √3 (tan x + 1) = 0
=> (tan x − √3) (tan x + 1) = 0
=> tan x = √3 or tan x = −1
=> tan x = tan π/3 or tan x = −tan π/4
=> tan x = tan π/3 or tan x = tan (−π/4)
=> x = nπ + π/3 or x = nπ − π/4, n ∈ Z
(vi) 3 cos2 θ − 2√3 sin θ cos θ − 3 sin2 θ = 0
Solution:
We have,
=> 3 cos2 θ − 2√3 sin θ cos θ − 3 sin2 θ = 0
=> √3 cos2 θ − 2 sin θ cos θ − √3 sin2 θ = 0
=> √3 cos2 θ + sin θ cos θ − 3 sin θ cos θ − √3 sin2 θ = 0
=> cos θ (√3cos θ + sin θ) − √3sin θ (√3cos θ + sin θ) = 0
=> (√3cos θ + sin θ) (cos θ − √3sin θ) = 0
=> tan θ = −√3 or tan θ = 1/√3
=> tan θ = − tan π/3 or tan θ = tan π/6
=> tan θ = tan (−π/3) or tan θ = tan π/6
=> θ = nπ − π/3 or θ = nπ + π/6, n ∈ Z
(vii) cos 4θ = cos 2θ
Solution:
We have,
=> cos 4θ = cos 2θ
=> cos 4θ − cos 2θ = 0
=> 2 sin 3θ sin θ = 0
=> sin 3θ = 0 or sin θ = 0
=> 3θ = nπ or θ = nπ
=> θ = nπ/3 or θ = nπ, n ∈ Z
Question 4. Solve the following equations:
(i) cos θ + cos 2θ + cos 3θ = 0
Solution:
We are given,
=> cos 2θ + (cos θ + cos 3θ) = 0
=> cos 2θ + 2cos 2θ cos θ = 0
=> cos 2θ (1 + 2cos θ) = 0
=> cos 2θ = 0 or cos θ = −1/2
=> 2θ = (2n+1)π/2 or cos θ = cos (π − π/3)
=> θ = (2n+1)π/4 or θ = 2nπ ± (2π/3), n ∈ Z
(ii) cos θ + cos 3θ − cos 2θ = 0
Solution:
We are given,
=> cos θ + cos 3θ − cos 2θ = 0
=> 2cos 2θ cos θ − cos 2θ = 0
=> cos 2θ (2cos θ − 1) = 0
=> cos 2θ = 0 or cos θ = 1/2
=> 2θ = (2n+1)π/2 or cos θ = cos π/3
=> θ = (2n+1)π/4 or θ = 2nπ ± (π/3), n ∈ Z
(iii) sin θ + sin 5θ = sin 3θ
Solution:
We are given,
=> sin θ + sin 5θ = sin 3θ
=> 2sin 3θ cos 2θ − sin 3θ = 0
=> sin 3θ (2cos 2θ − 1) = 0
=> sin 3θ = 0 or cos 2θ = 1/2
=> sin 3θ = 0 or cos 2θ = cos π/3
=> 3θ = nπ or 2θ = 2nπ ± (π/3)
=> θ = nπ/3 or θ = nπ ± (π/6), n ∈ Z
(iv) cos θ cos 2θ cos 3θ = 1/4
Solution:
We are given,
=> cos θ cos 2θ cos 3θ = 1/4
=> 2cos θ cos 3θ cos 2θ = 1/2
=> (cos 4θ + cos 2θ) cos 2θ = 1/2
=> (2cos2 2θ − 1 + cos 2θ) cos 2θ = 1/2
=> 2cos3 2θ − cos 2θ + cos2 2θ = 1/2
=> 4cos3 2θ − 2cos 2θ + 2cos2 2θ − 1 = 0
=> 2cos2 2θ (2cos 2θ + 1) − (2cos 2θ + 1) = 0
=> (2cos2 2θ − 1) (2cos 2θ + 1) = 0
=> 2cos2 2θ − 1 = 0 or 2cos 2θ + 1 = 0
=> cos 4θ = 0 or cos 2θ = −1/2
=> 4θ = (2n+1)π/2 or cos 2θ = cos 2π/3
=> θ = (2n+1)π/8 or 2θ = 2nπ ± (2π/3)
=> θ = (2n+1)π/8 or 2θ = nπ ± (π/3), n ∈ Z
(v) cos θ + sin θ = cos 2θ + sin 2θ
Solution:
We are given,
=> cos θ + sin θ = cos 2θ + sin 2θ
=> cos θ − cos 2θ = sin 2θ − sin θ
=> 2 sin 3θ/2 sin θ/2 = 2 cos 3θ/2 sin θ/2
=> 2 sin θ/2 (sin 3θ/2 − cos 3θ/2) = 0
=> sin θ/2 = 0 or (sin 3θ/2 − cos 3θ/2) = 0
=> θ/2 = nπ or tan 3θ/2 = 1
=> θ = 2nπ or tan 3θ/2 = tan π/4
=> θ = 2nπ or 3θ/2 = nπ ± π/4
=> θ = 2nπ or θ = 2nπ/3 ± π/6, n ∈ Z
(vi) sin θ + sin 2θ + sin 3θ = 0
Solution:
We are given,
=> (sin θ + sin 3θ) + sin 2θ = 0
=> 2sin 2θ cos θ + sin 2θ = 0
=> sin 2θ (2cos θ + 1) = 0
=> sin 2θ = 0 or 2cos θ + 1 = 0
=> 2θ = nπ or cos θ = −1/2
=> θ = nπ/2 or cos θ = cos 2π/3
=> θ = nπ/2 or θ = 2nπ ± (2π/3), n ∈ Z
(vii) sin x + sin 2x + sin 3x + sin 4x = 0
Solution:
We are given,
=> sin x + sin 2x + sin 3x + sin 4x = 0
=> (sin x + sin 3x) + (sin 2x + sin 4x) = 0
=> 2 sin 2x cos x + 2 sin 3x cos x = 0
=> 2cos x (sin 2x + sin 3x) = 0
=> (2cos x) (2sin 5x/2) (cos x/2) = 0
=> cos x = 0 or sin 5x/2 = 0 or cos x/2 = 0
=> x = (2n+1)π/2 or 5x/2 = nπ or cos x/2 = 0
=> x = (2n+1)π/2 or x = 2nπ/5 or x/2 = (2n+1)π/2
=> x = (2n+1)π/2 or x = 2nπ/5 or x = (2n+1)π, n ∈ Z
(viii) sin 3θ − sin θ = 4 cos2 θ − 2
Solution:
We are given,
=> sin 3θ − sin θ = 4 cos2 θ − 2
=> 2 cos 2θ sin θ = 2 (2cos2 θ − 1)
=> 2 cos 2θ sin θ = 2 cos 2θ
=> 2 cos 2θ (sin θ − 1) = 0
=> cos 2θ = 0 or (sin θ − 1) = 0
=> 2θ = (2n+1)π/2 or sin θ = 1
=> θ = (2n+1)π/4 or sin θ = sin π/2
=> θ = (2n+1)π/4 or θ = nπ + (−1)n (π/2), n ∈ Z
(ix) sin 2x− sin 4x + sin 6x = 0
Solution:
We are given,
=> sin 2x− sin 4x + sin 6x = 0
=> 2 sin 4x cos 2x − sin 4x = 0
=> 2 sin 4x (cos 2x − 1) = 0
=> sin 4x = 0 or cos 2x = 1/2
=> 4x = nπ or cos 2x = cos π/3
=> x = nπ/4 or 2x = 2nπ ± (π/3)
=> x = nπ/4 or x = nπ ± (π/6), n ∈ Z
Question 5. Solve the following equations:
(i) tan x + tan 2x + tan 3x = 0
Solution:
We are given,
=> tan x + tan 2x + tan 3x = 0
=> (tan x + tan 2x) + tan (2x+x) = 0
=> (tan x + tan 2x) + (tan x + tan 2x)/(1 − tan x tan 2x) = 0
=> (tan x + tan 2x) [1 + 1/(1 − tan x tan 2x)] = 0
=> (tan x + tan 2x) (2 − tan x tan 2x) = 0
=> (tan x + tan 2x) = 0 or (2 − tan x tan 2x) = 0
=> tan x = −tan 2x or tan x tan 2x = 2
=> tan x = tan (−2x) or tan x [2tan x/(1−tan2 x)] = 2
=> x = nπ − 2x or 2tan2 x = 2 − tan2 x
=> 3x = nπ or 4tan2 x = 2
=> x = nπ/3 or tan2 x = 1/2
=> x = nπ/3 or tan x = 1/√2 or tan x = −1/√2
=> x = nπ/3 or tan x = tan π/4 or tan x = tan (−π/4)
=> x = nπ/3 or x = nπ + π/4 or x = nπ − π/4
=> x = nπ/3 or x = nπ ± π/4, n ∈ Z
(ii) tan θ + tan 2θ = tan 3θ
Solution:
We are given,
=> tan θ + tan 2θ = tan 3θ
=> (tan θ + tan 2θ) − tan (2θ+θ) = 0
=> (tan θ + tan 2θ) − (tan θ + tan 2θ)/(1 − tan θ tan 2θ) = 0
=> (tan θ + tan 2θ) [1 − 1/(1 − tan θ tan 2θ)] = 0
=> (tan θ + tan 2θ) (−tan θ tan 2θ) = 0
=> (tan θ + tan 2θ) = 0 or −tan θ = 0 or tan 2θ = 0
=> tan θ = tan (−2θ) or tan θ = 0 or tan 2θ = 0
=> θ = nπ − 2θ or θ = nπ or 2θ = nπ
=> θ = nπ/3 or θ = nπ or θ = nπ/2, n ∈ Z
(iii) tan 3θ + tan θ = 2tan 2θ
Solution:
We are given,
=> tan 3θ + tan θ = 2tan 2θ
=> tan 3θ − tan 2θ = tan 2θ − tan θ
=> 2 sin2 θ sin 2θ = 0
=> sin θ = 0 or sin 2θ = 0
=> θ = nπ or θ = nπ/2, n ∈ Z
Question 6. Solve the following equations:
(i) sin θ + cos θ = √2
Solution:
We are given,
=> sin θ + cos θ = √2
=> (1/√2) sin θ + (1/√2) cos θ = 1
=> sin π/4 sin θ + cos π/4 cos θ = 1
=> cos (θ − π/4) = cos 0
=> θ − π/4 = 2nπ
=> θ = 2nπ + π/4
=> θ = (8n+1)π/4, n ∈ Z
(ii) √3 cos θ + sin θ = 1
Solution:
We are given,
=> √3 cos θ + sin θ = 1
=> (√3/2) cos θ + (1/2) sin θ = 1/2
=> cos π/6 cos θ + sin π/6 sin θ = 1/2
=> cos (θ − π/6) = cos π/3
=> θ − π/6 = 2nπ ± π/3
=> θ = 2nπ ± π/3 + π/6
=> θ = 2nπ + π/3 − π/6 or θ = 2nπ − π/3 + π/6
=> θ = 2nπ + π/2 or θ = 2nπ − π/6
=> θ = (4n+1)π/2 or θ = (12n−1)π/6, n ∈ Z
(iii) sin θ + cos θ = 1
Solution:
We are given,
=> sin θ + cos θ = 1
=> (1/√2) cos θ + (1/√2) sin θ = 1/√2
=> cos π/4 cos θ + sin π/4 sin θ = 1/√2
=> cos (θ − π/4) = cos π/4
=> θ − π/4 = 2nπ ± π/4
=> θ = 2nπ ± π/4 + π/4
=> θ = 2nπ + π/2 or θ = 2nπ, n ∈ Z
(iv) cosec θ = 1 + cot θ
Solution:
We are given,
=> cosec θ = 1 + cot θ
=> 1/sin θ = 1 + cos θ/sin θ
=> sin θ + cos θ = 1
=> (1/√2) cos θ + (1/√2) sin θ = 1/√2
=> cos π/4 cos θ + sin π/4 sin θ = 1/√2
=> cos (θ − π/4) = cos π/4
=> θ − π/4 = 2nπ ± π/4
=> θ = 2nπ ± π/4 + π/4
=> θ = 2nπ + π/2 or θ = 2nπ, n ∈ Z
(v) (√3 − 1) cos θ + (√3 + 1) sin θ = 2
Solution:
We are given,
=> (√3 − 1) cos θ + (√3 + 1) sin θ = 2
=> (√3 − 1) cos θ/2√2 + (√3 + 1) sin θ/2√2 = 2
=> sin (θ + tan-1 (√3 − 1)/(√3 + 1)) = sin π/4
=> θ = 2nπ + π/3 or θ = 2nπ − π/6, n ∈ Z
Question 7. Solve the following equations:
(i) cot x + tan x = 2
Solution:
We are given,
=> cot x + tan x = 2
=> cos x/sin x + sin x/cos x = 2
=> (cos2 x + sin2 x)/sin x cos x = 2
=> 2 sin x cos x = 1
=> sin 2x = 1
=> sin 2x = sin π/2
=> 2x = (2n+1)π/2
=> x = (2n+1)π/4, n ∈ Z
(ii) 2 sin2 θ = 3 cos θ, 0 ≤ θ ≤ 2π
Solution:
We are given,
=> 2 sin2 θ = 3 cos θ
=> 2 (1 − cos2 θ) − 3 cos θ = 0
=> 2 cos2 θ + 3 cos θ − 2 = 0
=> 2 cos2 θ + 4 cos θ − cos θ − 2 = 0
=> 2 cos θ (cos + 2) − (cos θ + 2) = 0
=> (2 cos θ − 1) (cos θ + 2) = 0
=> cos θ = 1/2 or cos θ = −2
Ignoring cos θ =−2 as −1 ≤ cos θ ≤ 1. So, we have, cos θ = 1/2.
=> cos θ = cos π/3
=> θ = 2nπ ± π/3, n ∈ Z
(iii) sec x cos 5x +1 = 0, 0 ≤ x ≤ π/2
Solution:
We are given,
=> sec x cos 5x +1 = 0
=> (cos5x + cos x)/cos x = 0
=> 2 cos 3x cos 2x = 0
=> cos 3x = 0 or cos 2x = 0
=> 3x = (2n+1)π/2 or 2x = (2n+1)π/2
=> x = (2n+1)π/6 or x = (2n+1)π/4, n ∈ Z
(iv) 5 cos2 θ + 7 sin2 θ − 6 = 0
Solution:
We are given,
=> 5 cos2 θ + 7 sin2 θ − 6 = 0
=> 5 (1 − sin2 θ) + 7 sin2 θ − 6 = 0
=> 2 sin2 θ+ 5 − 6 = 0
=> 2 sin2 θ = 1
=> sin θ = ±(1/√2)
=> θ = nπ ± π/4, n ∈ Z
(v) sin x − 3 sin 2x + sin 3x = cos x − 3 cos 2x + cos 3x
Solution:
We are given,
=> sin x − 3 sin 2x + sin 3x = cos x − 3 cos 2x + cos 3x
=> (sin x + sin 3x) − 3 sin 2x − (cos x + cos 3x) + 3 cos 2x = 0
=> 2 sin 2x cos x − 3 sin 2x − 2 cos 2x cos x + 3 cos 2x = 0
=> sin 2x (2 cos x − 3) − cos 2x (2 cos x − 3) = 0
=> (sin 2x − cos 2x) (2 cos x − 3) = 0
=> sin 2x = cos 2x or cos x = 3/2
Ignoring cos x = 3/2 as −1 ≤ cos x ≤ 1. So, we have, sin 2x = cos 2x.
=> tan 2x = 1
=> tan 2x = tan π/4
=> 2x = nπ + π/4
=> x = nπ/2 + π/8, n ∈ Z
Conclusion
Exercise 11.1 in Chapter 11 serves as a stepping stone for the mastering trigonometric equations. By working through these problems students gain essential skills in solving equations involving the trigonometric functions which are crucial for the higher-level mathematics. The concepts learned here are not only important for the exams but also for the practical applications in the various scientific fields.
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