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TRIGONOMETRIC EQUATION
Ex.16 If sin 5x + sin 3x + sinx = 0 and 0 ≤ x ≤ π/2, then x is equal to
[1] π/12
[2] π/6
[3] π/4
Sol. sin 5x + sinx = – sin 3x
⇒ 2 sin 3x cos 2x + sin 3x = 0
⇒ sin 3x (2 cos 2x + 1) = 0
⇒ sin 3x = 0, cos 2x = – 1/2
⇒ x = nπ, x = nπ ± (π/3)
So x = π/3
[4] π/3
Ans. [4]
Ex.17 The number of solutions of equation, sin 5x cos 3x = sin 6x cos 2x, in the interval [0, π] are
[1] 3
[2] 4
[3] 5
[4] 6
Sol. The given equation can be written as
1
1
(sin 8x + sin 2x) =
(sin 8x + sin 4x)
2
2
or, sin 2x – sin 4x ⇒ – 2 sin xcos 3x = 0
Hence sin x = 0 or cos 3x = 0. That is, x = nπ(n ∈ Ι), or 3x = kπ +
x ∈ [0, π], then given equation is satisfied if x = 0, π,
π
(k ∈ Ι). Therefore, since
2
π π 5π
, or
6 2
6
Ans. [3]
Ex.18 Solve the following system of equation sinx + cosy =1 , cos2x – cos2y = 1
Sol.
Given sinx + cosy = 1
......(i)
and (1 – 2sin2x) – (2cos2y –1) = 1 ⇒ sin2x + cos2y = 1/2
Put sin x = u and cosy = v in (i) and (ii) u + v = 1 ⇒ u2 + v2 = 1/ 2
solving above equations u =
π
1
1
1
and v =
, n ∈I
⇒ sin x =
⇒ x = nπ + (–1)n
6
2
2
2
π
1
⇒ y = 2mπ ± , m ∈ I
⇒ ∴ the given equation have solution
3
2
π
π
x = nπ + (–1)n
, n ∈ I and y = 2mπ ± , m ∈ I
6
3
cosy =
.....(ii)
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