find-solution-4-sin-x-1-4-1-2-2-2-sin-x-1-0-in-x-0-2pi-

Question Number 86701 by john santu last updated on 30/Mar/20

find solution

4^{\sin x - \frac{1}{4}} - \frac{1}{2 + \sqrt{2}} . 2^{\sin x} - 1 = 0

in x \in [0, 2 π]

Commented by john santu last updated on 30/Mar/20

4^{\sin x} . 4^{- \frac{1}{4}} - (\frac{2 - \sqrt{2}}{2}) . 2^{\sin x} - 1 = 0

\sqrt{2} 4^{\sin x} - (2 - \sqrt{2}) . 2^{\sin x} - 2 = 0

\sqrt{2} . 2^{\sin x} (2^{\sin x} + 1) - 2 (2^{\sin x} + 1) = 0

(2^{\sin x} + 1) (\sqrt{2} . 2^{\sin x} - 2) = 0

\Rightarrow \sqrt{2} . 2^{\sin x} - 2 = 0, 2^{\sin x} = 2^{0.5}

\sin x = 0.5 \Rightarrow {\begin{cases} x = \frac{π}{6} \\ x = \frac{5 π}{6} \end{cases}