Question

If $\mathbf{\sin}^{\mathbf{10}}\mathbf{x}\mathbf{+}\mathbf{\cos}^{\mathbf{10}}\mathbf{x}\mathbf{=}\frac{\mathbf{29}}{\mathbf{16}}\mathbf{\cos}^{\mathbf{4}}\mathbf{2}\mathbf{x.}$ then $x =$

• A. $x = \frac{n\pi}{4}$
• B. $x = \frac{n\pi}{4} + \frac{\pi}{8}$
• C. $x = \frac{n\pi}{4} + \frac{\pi}{3}$
• D. None of these

Answer 1

Answer: (B)

$x = \frac{n\pi}{4} + \frac{\pi}{8}$

Answer 2

Using half-angle formulae we can represent the given equation in the form,

$\left( \frac{1 - \cos 2x}{2} \right)^{5} + \left( \frac{1 + \cos 2x}{2} \right)^{5} = \frac{29}{16}\cos^{4}2x$ Put $\cos 2 x = t$ $\left( \frac{1 - t}{2} \right)^{5} + \left( \frac{1 + t}{2} \right)^{5} = \frac{29}{16}t^{4} \Rightarrow$ $24t^{4} - 10t^{2} - 1 = 0$ whose only

real root is, $t^{2} = \frac{1}{2}.$

$\therefore$ $\cos^{2}2x = \frac{1}{2}$ $\Rightarrow$ $1 + \cos 4x = 1$ $\Rightarrow$ $\cos 4x = 0$

$\Rightarrow 4x = (2n + 1)\frac{\pi}{2}$ $\Rightarrow$ $x = \frac{n\pi}{4} + \frac{\pi}{8}$; $n \in$I