Download Page 1 1 TRIGONOMETRIC EQUATIONS PREVIOUS EAMCET

TRIGONOMETRIC EQUATIONS
PREVIOUS EAMCET BITS
1.
If 3cos x ≠ 2,sin x , then the general solution of sin 2 x − cos 2x = 2 − sin 2x is x =
[EAMCET 2009]
1) nπ + ( −1)
n
π
,n∈Z
2
2)
π
3) ( 4n ± 1) , n ∈ Z
2
Ans: 3
nπ
,n∈Z
2
4) ( 2n − 1) π, n ∈ Z
Sol. sin 2 x − (1 − 2sin 2 x ) = 2 − 2sin x cos x
⇒ 3sin 2 x + 2sin x cos x − 3 = 0
⇒ 3sin 2 x + 2sin x cos x − 3 ( sin 2 x + cos 2 x ) = 0
⇒ cos x ( 2sin x − 3cos x ) = 0
⇒ cos x = 0 (∵ 2sin x ≠ 3cos x )
⇒ x = ( 4n ± 1) π / 2, n ∈ Z
2.
{x ∈ R : cos 2x + 2 cos
2
x − 2 = 0} =
[EAMCET 2008]
π
π
π
⎧
⎫
⎧
⎫ ⎧
⎫
1) ⎨2nπ + ; n ∈ Z⎬ 2) ⎨nπ ± ; n ∈ Z⎬
3) ⎨nπ ± ; n ∈ Z⎬
3
6
3
⎩
⎭ ⎩
⎭
⎩
⎭
Ans: 2
Sol. cos 2x + 2 cos 2 x − 2 = 0 ⇒ 2 cos 2 x − 1 + 2 cos 2 x − 2 = 0
3
π
⎛π⎞
⇒ 4 cos 2 x = 3 ⇒ cos 2 x = = cos 2 ⎜ ⎟ ⇒ x = nπ ±
4
6
⎝6⎠
(
)
1 ⎞
1
⎛
2 + 1 ⎜ cos x −
⎟ , cos x ≠ 2 ⇒ x ∈
2⎠
⎝
π
π
⎧
⎫
⎧
⎫
2) ⎨2nπ ± : n ∈ Z ⎬
1) ⎨2nπ ± : n ∈ Z ⎬
3
6
⎩
⎭
⎩
⎭
π
π
⎧
⎫
⎧
⎫
3) ⎨2nπ ± : n ∈ Z⎬
4) ⎨2nπ ± : n ∈ Z⎬
2
4
⎩
⎭
⎩
⎭
Ans: 4
Sol. 2 2 cos 2 x − 2 + 2 cos x + 1 = 0
3.
cos 2x =
(
)
2 2 cos x − 2 cos x − 2 cos x + 1 = 0
2
( 2 cos x − 1) (
cos x ≠
1
;
2
)
2 cos x − 1 = 0
cos x =
1
2
1
π
⎧
⎫
4) ⎨2nπ − ; n ∈ Z⎬
3
⎩
⎭
[EAMCET 2005]
Trigonometric Equations
4.
π
⎧
⎫
∴ x ∈ ⎨ 2nπ ± / n ∈ Z ⎬
4
⎩
⎭
The solution set of ( 5 + 4 cos θ )( 2 cos θ + 1) = 0 in the interval [0, 2π] is
⎧ π 2π ⎫
⎧ 2 π 5π ⎫
2) ⎨ , ⎬
1) ⎨ , ⎬
⎩3 3 ⎭
⎩3 3⎭
Ans: 3
Sol. ( 5 + 4 cos θ )( 2 cos θ + 1) = 0
2π 4π
−1
,
⇒θ=
2
3 3
5.
The equation 3 sin x + cos x = 4 has
1) only one solution
3) infinitely many solutions
Ans: 4
Sol. The max. value of 3 sinx + cosx is ‘2’.
∴ 3 sinx + cosx never equal to ‘4’.
⎧ 2π 4 π ⎫
3) ⎨ , ⎬
⎩3 3⎭
[EAMCET 2003]
⎧ 2 π 5π ⎫
4) ⎨ , ⎬
⎩3 3⎭
cos θ =
6.
[EAMCET 2001]
2) two solutions
4) no solutions
If tan θ + sec θ = 3 , then the principal value of θ +
1) π/4
Ans: 2
Sol. tan θ + sec θ = 3
2) π/3
3) 2π/3
3 cos θ − sin θ = 1
π⎞ 1
π⎞ π
⎛
⎛
∴⎜θ+ ⎟ =
cos ⎜ θ + ⎟ =
6⎠ 2
6⎠ 3
⎝
⎝
› ››
2
π
is
6
[EAMCET 2000]
4) 3π/4