Q. 7 B5.0( 2 Votes )

# Solve the followi

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ny, where n Z.

• cos x = cos y, implies x = 2nπ ± y, where n Z.

• tan x = tan y, implies x = nπ + y, where n Z.

given,

2 sin2 x = 3 cos x , 0 ≤ x ≤ 2π

2 ( 1 – cos2 x) = 3 cos x

2 cos2 x + 3cos x – 2 = 0

2 cos2 x + 4 cos x – cos x – 2 = 0

2 cos x(cos x + 2) – 1(cos x + 2) = 0

(2cos x – 1)(cos x + 2) = 0

cos x = �

or cos x = -2 { as cos x lies between -1 and 1 so this value is rejected }

cos x = � = cos π/3

If cos x = cos y, implies x = 2nπ ± y, where n Z

x = 2nπ ± π/3

But, 0 ≤ x ≤ 2π

x = π/3 and x = 2π - π/3 = 5π/3 ….ans

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