Q. 3 C5.0( 1 Vote )

# 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)n y, 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,

2sin2 x +√3 cos x + 1 = 0

As the equation is of 2nd degree, so we need to solve a quadratic equation.

First we will substitute trigonometric ratio with some variable k and we will solve for k

As, sin2 x = 1 – cos2 x

we have,

Let, cos x = k

2k2 - √3 k – 3 = 0

2k2 -2√3 k + √3 k – 3 = 0

2k(k – √3) +√3(k – √3) = 0

(2k + √3)(k - √3) = 0

k = √3 or k = -√3/2

cos x = √3 or cos x = -√3/2

As cos x lies between -1 and 1

cos x can’t be √3

So we ignore that value.

cos x = -√3/2

cos x = cos 150° = cos 5π/6

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

On comparing our equation with standard form, we have

y = 5π/6

where n ϵ Z ..ans

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