Trigonometric equations

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The number of solutions of $(\sin 2 x + \cos 2 x)^{1 + \sin 4 x} = 2 in$ $[- π, π]$ is equal to

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$\begin{array}{l} (\sin 2 x + \cos 2 x)^{(\sin 2 x + \cos 2 x)^{2}} = 2 \\ \Rightarrow \sin 2 x + \cos 2 x = \sqrt{2} \\ \Rightarrow \sin (2 x + \frac{π}{4}) = 1 \\ \Rightarrow 2 x + \frac{π}{4} = 2 nπ + \frac{π}{2} \Rightarrow x = nπ + \frac{π}{8} \end{array}$
for, $x \in [- π, π], put n = - 1, 0, 1$
we get, $x = - \frac{7 π}{8}, \frac{π}{8}$

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Trigonometric equations

Remember concepts with our Masterclasses.

The number of solutions of (sin⁡2x+cos⁡2x)1+sin⁡4x=2 in [−π,π] is equal to

(sin⁡2x+cos⁡2x)(sin⁡2x+cos⁡2x)2=2⇒sin⁡2x+cos⁡2x=2⇒sin⁡2x+π4=1⇒2x+π4=2nπ+π2⇒x=nπ+π8for, x∈[−π,π], put n=−1,0,1we get, x=−7π8,π8

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The number of solutions of $(\sin 2 x + \cos 2 x)^{1 + \sin 4 x} = 2 in$ $[- π, π]$ is equal to