First slide

Trigonometric equations

Question

The general solution of the equation $\frac{1 + \sin x + .... + \sin^{n} x + .....}{1 - \sin x + .... + {(- 1)}^{n} \sin^{n} x + ....} = \frac{1 - \cos 2 x}{1 + \cos 2 x}$ , $x \neq (2 n + 1) π / 2, n \in I$ is

Difficult

A

${(- 1)}^{n} (π / 3) + n π$
B

${(- 1)}^{n} (π / 6) + n π$
C

${(- 1)}^{n + 1} (π / 6) + n π$
D

${(- 1)}^{n - 1} (π / 3) + n π$

Solution

The equation $\frac{1 + \sin x + .... + \sin^{n} x + .....}{1 - \sin x + .... + {(- 1)}^{n} \sin^{n} x + ....}$ $= \frac{1 - \cos 2 x}{1 + \cos 2 x}$
$\Rightarrow \frac{1 + \sin x}{1 - \sin x} = \frac{2 \sin^{2} x}{2 \cos^{2} x}$
$\Rightarrow \frac{{(1 + \sin x)}^{2}}{\cos^{2} x} = \frac{\sin^{2} x}{\cos^{2} x}$
$\Rightarrow {(1 + \sin x)}^{2} = \sin^{2} x$
$\Rightarrow 1 + 2 \sin x = 0$
$\Rightarrow \sin x = - \frac{1}{2} = \sin (- \frac{π}{6})$
$x = n π + {(- 1)}^{n + 1} \frac{π}{6}$

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