The solution of the equation $[\sin x+\cos x]^{1+\sin 2 x}=2$, where $-\pi \leq x \leq \pi$ is
The solution of the equation $[\sin x+\cos x]^{1+\sin 2 x}=2$, where $-\pi \leq x \leq \pi$ is
- $\frac{\pi}{2}$
- $\pi$
- $\frac{\pi}{4}$
- $\frac{3 \pi}{4}$
Solution
$\begin{aligned} & \text { }(\sin x+\cos x)^{1+\sin 2 x}=2 \text { where }-\pi \leq x \leq \pi \\ & (\sin x+\cos x)^{(\sin x+\cos x)^2}=2\left[\because(\sin x+\cos x)^2\right. \\ & =\sin ^2 x+\cos ^2 x+2 \sin \cos x \\ & =1+\sin 2 x] \\ & \because \sqrt{a^2+b^2} \leq a \sin x+b \cos x \leq \sqrt{a^2+b^2} \\ & \therefore-\sqrt{1^2+1^2} \leq \sin x+\cos x \leq \sqrt{1^2+1^2} \\ & \Rightarrow \quad-\sqrt{2} \leq \sin x+\cos x \leq \sqrt{2} \\ & \text { For }-\pi \leq x \leq \pi \\ & \because \sin x+\cos x=-\sqrt{2} \text { at } x=-\frac{\pi}{4} \\ & \end{aligned}$
$
\sin x+\cos x=\sqrt{2} \text { at } x=\pi / 4
$
Now, at $x=\pi / 4$
We get
$
\begin{aligned}
(\sqrt{2})^{(\sqrt{2})^2} & =2 \\
& =(\sqrt{2})^2=2=2=2
\end{aligned}
$
$\therefore$ It is a solution.
At $x=-\frac{\pi}{4}$
We get
$
\begin{aligned}
(-\sqrt{2})^{(-\sqrt{2})^2} & =2 \\
(-\sqrt{2})^2 & =2 \Rightarrow 2=2
\end{aligned}
$
It is also a solution
Asked in: AP EAMCET 2021 (24 Aug Shift 1)
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