Question

Solve the trigonometric equation given by
( sin2x - cosx ) / ( cos2x + sinx - 1 ) = 0 → for 0 ≤ x ≤ 2π

Answer 1

${\huge{\bold{\underline{Answer:-}}}}$

Replacing cos2x by 1 − sin2x, we get a quadratic in sin of the form

4sin2x + 4sinx −3 = 0

⇒ (2sinx + 3) (2sinx −1) = 0

Now

sin x ≠ − 3/2 since |sin x| ≤ 1

Therefore, sin x = 1/2

The principal solution is x = π/6.

The general solution is x = nπ +(-1)nπ/6.

Answer 2

SOLUTION

We know that

• sin2x = 2 sinx cosx and
• cos2x = 1 - 2 sin2x

NOW

( sin2x - cosx ) / ( cos2x + sinx - 1 ) = 0

( 2 sinx cosx - cosx ) / ( 1 - 2 sin2x + sinx - 1) = 0

cosx( 2 sinx - 1 ) / [ - sinx ( 2 sinx - 1 ) ] = 0

• Divide numerator and denominator by 2 sin(x) - 1 to simplify & 2 sin(x) - 1 ≠ 0.

- cosx / sinx = 0

-cotx = 0

Value of x = π/2 & 3π/2

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