જો 5(tan^2 x - cos^2 x) = 2cos 2x + 9 હોય,તો | vedclass

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Question

(A) આપેલ સમીકરણ: $5(	an^2 x - \cos^2 x) = 2\cos 2x + 9$$\cos 2x = 2\cos^2 x - 1$ નો ઉપયોગ કરતા:$5	an^2 x - 5\cos^2 x = 2(2\cos^2 x - 1) + 9$$5	an^2

Vedclass · Accepted Answer

$-\frac{7}{9}$

Vedclass · Answer

(A) આપેલ સમીકરણ: $5(	an^2 x - \cos^2 x) = 2\cos 2x + 9$$\cos 2x = 2\cos^2 x - 1$ નો ઉપયોગ કરતા:$5	an^2 x - 5\cos^2 x = 2(2\cos^2 x - 1) + 9$$5	an^2 x - 5\cos^2 x = 4\cos^2 x - 2 + 9$$5	an^2 x = 9\cos^2 x + 7$$	an^2 x = \frac{1}{\cos^2 x} - 1$ અને $t = \cos^2 x$ લેતા:$5(\frac{1}{t} - 1) = 9t + 7$$9t^2 + 12t - 5 = 0$$(3t - 1)(3t + 5) = 0$$t = \cos^2 x$ ધન હોવાથી,$t = \frac{1}{3}$.હવે,$\cos 2x = 2(\frac{1}{3}) - 1 = -\frac{1}{3}$.તેથી,$\cos 4x = 2(-\frac{1}{3})^2 - 1 = \frac{2}{9} - 1 = -\frac{7}{9}$.

જો $5(\tan^2 x - \cos^2 x) = 2\cos 2x + 9$ હોય,તો $\cos 4x$ ની કિંમત શોધો.

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