find solution 4^(sin x −(1/4)) − (1/(2+(√2))) .2^(sin x) −1 = 0 in x ∈[ 0,2π ]


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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
$4^(sin x) .4^(−(1/4)) −(((2−(√2))/2)).2^(sin x) −1 = 0 (√2) 4^(sin x) −(2−(√2)).2^(sin x) −2 = 0 (√2).2^(sin x) (2^(sin x) +1)−2(2^(sin x) +1) =0 (2^(sin x) +1)((√2).2^(sin x) −2) =0 ⇒(√2) .2^(sin x) −2 = 0, 2^(sin x) = 2^(0.5) sin x = 0.5 ⇒ { ((x = (π/6))),((x = ((5π)/6))) :}$
$4^{\sin x} . 4^{- \frac{1}{4}} - (\frac{2 - \sqrt{2}}{2}) . 2^{\sin x} - 1 = 0$ $\sqrt{2} 4^{\sin x} - (2 - \sqrt{2}) . 2^{\sin x} - 2 = 0$ $\sqrt{2} . 2^{\sin x} (2^{\sin x} + 1) - 2 (2^{\sin x} + 1) = 0$ $(2^{\sin x} + 1) (\sqrt{2} . 2^{\sin x} - 2) = 0$ $\Rightarrow \sqrt{2} . 2^{\sin x} - 2 = 0, 2^{\sin x} = 2^{0.5}$ $\sin x = 0.5 \Rightarrow {\begin{cases} x = \frac{π}{6} \\ x = \frac{5 π}{6} \end{cases}$
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