જો cos left(frac{pi}{4}-xright) cos 2 x+sin x | vedclass

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Question

(D) આપેલ સમીકરણ: $\cos \left(\frac{\pi}{4}-x\right) \cos 2 x+\sin x \sin 2 x \sec x = \cos x \sin 2 x \sec x+\cos \left(\frac{\pi}{4}+x\right) \cos 2

Vedclass · Accepted Answer

$\sqrt{2}$

Vedclass · Answer

(D) આપેલ સમીકરણ: $\cos \left(\frac{\pi}{4}-xight) \cos 2 x+\sin x \sin 2 x \sec x = \cos x \sin 2 x \sec x+\cos \left(\frac{\pi}{4}+xight) \cos 2 x$ પદોને ગોઠવતા: $\cos 2 x \left[ \cos \left(\frac{\pi}{4}-xight) - \cos \left(\frac{\pi}{4}+xight) ight] = \sin 2 x \sec x (\cos x - \sin x)$ સૂત્ર $\cos(A-B) - \cos(A+B) = 2 \sin A \sin B$ નો ઉપયોગ કરતા: $\cos 2 x \left[ 2 \sin \frac{\pi}{4} \sin x ight] = \sin 2 x \sec x (\cos x - \sin x)$ $\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$ અને $\sin 2 x = 2 \sin x \cos x$ હોવાથી: $\cos 2 x \left( \frac{2}{\sqrt{2}} \sin x ight) = (2 \sin x \cos x) \sec x (\cos x - \sin x)$ $\sqrt{2} \cos 2 x \sin x = 2 \sin x (\cos x - \sin x)$ $\sin x 
eq 0$ ધારીને,$\sin x$ વડે ભાગતા: $\sqrt{2} (\cos^2 x - \sin^2 x) = 2 (\cos x - \sin x)$ $\sqrt{2} (\cos x - \sin x)(\cos x + \sin x) = 2 (\cos x - \sin x)$ જો $\cos x - \sin x 
eq 0$ હોય,તો $\sqrt{2} (\cos x + \sin x) = 2$,એટલે કે $\cos x + \sin x = \sqrt{2}$. $\sqrt{2}$ વડે ભાગતા,$\frac{1}{\sqrt{2}} \cos x + \frac{1}{\sqrt{2}} \sin x = 1$,જે $\cos \left(x - \frac{\pi}{4}ight) = 1$ છે. તેથી,$x - \frac{\pi}{4} = 0$,એટલે કે $x = \frac{\pi}{4}$. તેથી $\sec x = \sec \frac{\pi}{4} = \sqrt{2}$.

જો $\cos \left(\frac{\pi}{4}-x\right) \cos 2 x+\sin x \sin 2 x \sec x = \cos x \sin 2 x \sec x+\cos \left(\frac{\pi}{4}+x\right) \cos 2 x$ હોય,તો $\sec x$ ની શક્ય કિંમત કઈ છે?

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