First slide

Trigonometric equations

Question

$T h e g e n e r a l s o l u t i o n o f t h e e q u a t i o n$ $\frac{1 - \sin x + ... + {(- 1)}^{n} \sin^{n} x + ......}{1 + \sin x + ... + \sin^{n} x + ......} = \frac{1 - \cos 2 x}{1 + \cos 2 x}$ $i s$

Moderate

A

${(- 1)}^{n} (\frac{π}{3}) + n π, \forall n \in I$
B

${(- 1)}^{n} (\frac{π}{6}) + n π, \forall n \in I$
C

${(- 1)}^{n + 1} (\frac{π}{6}) + n π, \forall n \in I$
D

${(- 1)}^{n - 1} (\frac{π}{3}) + n π, \forall n \in I$

Solution

$\frac{1 - \sin x + ... + {(- 1)}^{n} \sin^{n} x + ......}{1 + \sin x + ... + \sin^{n} x + ......} = \frac{1 - \cos 2 x}{1 + \cos 2 x}$
$\Rightarrow \frac{1}{1 + \sin x} . \frac{1 - \sin x}{1} = \frac{2 \sin^{2} x}{2 \cos^{2} x}$
$\begin{array}{l} \Rightarrow 2 \sin^{2} x + \sin x - 1 = 0 \\ \Rightarrow \sin x = \frac{- 1 \pm \sqrt{1 + 8}}{4} = \frac{- 1 \pm 3}{4} \\ \Rightarrow \sin x = - 1 o r \sin x = \frac{1}{2} \\ S i n c e \sin x \neq - 1, w e h a v e \sin x = \frac{1}{2} = \sin (\frac{π}{6}) \\ ∴ x = n π + {(- 1)}^{n} \frac{π}{6} \end{array}$

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