cosx - secx = 3/2 হলে, cos⁴x+sin⁴x এর মান ???ত?

Explanation: Solve: \( \cos x - \frac{1}{\cos x} = \frac{3}{2} \implies \cos^2 x - 1 = \frac{3}{2} \cos x \) \(\implies 2 \cos^2 x - 3 \cos x - 2 = 0 \implies 2 \cos x (\cos x - 2) + (\cos x - 2) = 0\) \(\implies (\cos x - 2)(2 \cos x + 1) = 0\) \(\implies \cos x \neq 2 \, (\text{গ্রহণযোগ্য নয়}), \cos x = -\frac{1}{2} = \cos \frac{2\pi}{3} \implies x = 120^\circ\) \(\therefore \cos^4 x + \sin^4 x = (\cos 120^\circ)^4 + (\sin 120^\circ)^4 = \left(-\frac{1}{2}\right)^4 + \left(\frac{\sqrt{3}}{2}\right)^4 = \frac{1}{16} + \frac{9}{16} = \frac{10}{16} = \frac{5}{8}\)

Another Explanation (5): ```html

দেওয়া আছে, \( \cos x - \sec x = \frac{3}{2} \)

আমরা জানি, \( \sec x = \frac{1}{\cos x} \)

সুতরাং, \( \cos x - \frac{1}{\cos x} = \frac{3}{2} \)

বা, \( \frac{\cos^2 x - 1}{\cos x} = \frac{3}{2} \)

বা, \( 2(\cos^2 x - 1) = 3\cos x \)

বা, \( 2\cos^2 x - 3\cos x - 2 = 0 \)

বা, \( 2\cos^2 x - 4\cos x + \cos x - 2 = 0 \)

বা, \( 2\cos x (\cos x - 2) + 1(\cos x - 2) = 0 \)

বা, \( (\cos x - 2)(2\cos x + 1) = 0 \)

যেহেতু, \( \cos x \neq 2 \), সুতরাং \( 2\cos x + 1 = 0 \)

অতএব, \( \cos x = -\frac{1}{2} \)

এখন, \( \cos^4 x + \sin^4 x = \cos^4 x + (1 - \cos^2 x)^2 \)

\( = \cos^4 x + (1 - 2\cos^2 x + \cos^4 x) \)

\( = 2\cos^4 x - 2\cos^2 x + 1 \)

\( = 2(-\frac{1}{2})^4 - 2(-\frac{1}{2})^2 + 1 \)

\( = 2(\frac{1}{16}) - 2(\frac{1}{4}) + 1 \)

\( = \frac{1}{8} - \frac{1}{2} + 1 \)

\( = \frac{1 - 4 + 8}{8} \)

\( = \frac{5}{8} \)

সুতরাং, \( \cos^4 x + \sin^4 x = \frac{5}{8} \) 🥳

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