જો sin^4,,alpha + 4,cos^4,,beta + 2 = 4sqrt 2 | vedclass

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Question

$\mathrm{AM} \geq \mathrm{GM}$

$\frac{\sin ^{4} \alpha+4 \cos ^{4} \beta+1+1}{4} \geq\left(\sin ^{4} \alpha \cdot 4 \cos ^{4} \beta .1 .1\right)^{\

Vedclass · Accepted Answer

$-\sqrt 2$

Vedclass · Answer

$\mathrm{AM} \geq \mathrm{GM}$

$\frac{\sin ^{4} \alpha+4 \cos ^{4} \beta+1+1}{4} \geq\left(\sin ^{4} \alpha \cdot 4 \cos ^{4} \beta .1 .1\right)^{\frac{1}{4}}$

$\sin ^{4} \alpha+4 \cos ^{2} \beta+$ $2 \geq 4 \sqrt{2} \sin \alpha \cos \beta$ given that $\sin ^{4} \alpha+4 \cos ^{4} \beta+2$ $=4 \sqrt{2} \sin \alpha \cos \beta$

$\Rightarrow \mathrm{AM}=\mathrm{GM} \Rightarrow \sin ^{4} \alpha=1=4 \cos ^{4} \beta$

$\sin \alpha=\pm 1, \cos \beta=\pm \frac{1}{\sqrt{2}},$ As $\alpha, \beta \in[0, \pi]$

$\Rightarrow \sin \alpha=1, \cos \beta=\pm \frac{1}{\sqrt{2}}$

$\Rightarrow \sin \beta=\frac{1}{\sqrt{2}}$ as $\beta \in[0, \pi]$

$\cos (\alpha+\beta)-\cos (\alpha-\beta)=-2 \sin \alpha \sin \beta$

$=-\sqrt{2}$

જો $sin^4\,\,\alpha + 4\,cos^4\,\,\beta + 2 = 4\sqrt 2\,\,sin\,\alpha \,cos\,\beta ;$ $\alpha \,,\,\beta \, \in \,[0,\pi ],$ તો $cos( \alpha + \beta)$ = ......

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આપેલ સમીકરણના વ્યાપક ઉકેલ શોધો : $\sin 2 x+\cos x=0$

$2 \cos ^{2} x+3 \sin x=0$ ઉકેલો.

જો ${\tan ^2}\theta - (1 + \sqrt 3 )\tan \theta + \sqrt 3 = 0$, તો $\theta $ નો વ્યાપક ઉકેલ મેળવો.

જો $P = \left\{ {\theta :\sin \,\theta - \cos \,\theta = \sqrt 2 \,\cos \,\theta } \right\}$ અને $Q = \left\{ {\theta :\sin \,\theta + \cos \,\theta = \sqrt {2\,} \sin \,\theta } \right\}$ બે ગણ હોય તો