1-sin-pi-7-3-cos-2x-sin-pi-14-cos-pi-14-10-sin-x-find-solution-

Question Number 78785 by jagoll last updated on 20/Jan/20

{(1 + \sin \frac{π}{7})}^{3 - \cos 2 x} = {(\sin \frac{π}{14} + \cos \frac{π}{14})}^{10 \sin x}

find solution

Answered by john santu last updated on 20/Jan/20

c o n s i d e r 1 + \sin \frac{π}{7} = {(\sin \frac{π}{14} + \cos \frac{π}{14})}^{2}

\Rightarrow {(\sin \frac{π}{14} + \cos \frac{π}{14})}^{6 - 2 \cos 2 x} = {(\sin \frac{π}{14} + \cos \frac{π}{14})}^{10 \sin x}

\Rightarrow 6 - 2 \cos 2 x = 10 \sin x

- 2 (1 - 2 \sin^{2} x) + 6 - 10 \sin x = 0

4 \sin^{2} x - 10 \sin x + 4 = 0

2 \sin^{2} x - 5 \sin x + 2 = 0

(2 \sin x - 1) (\sin x - 2) = 0

\sin x = \frac{1}{2} {\begin{cases} x = \frac{π}{6} + 2 n π \\ x = \frac{5 π}{6} + 2 n π \end{cases}

Commented by jagoll last updated on 20/Jan/20

thanks sir

Facebook Tweet Pin