The number of distinct solutions of the equation 5/4cos^2⁡ 2x+cos^4⁡x+sin^4⁡x+cos^6⁡x+sin^6⁡x=2 in the interval [0, 2π] is

Correct answer: 8. 5/4cos^2⁡2x+sin^2⁡x+cos^2⁡x2+sin^2⁡x+cos^2⁡x3-5sin^2⁡x.cos^2⁡x=2 ⇒5/41-sin^2⁡2x-5/4sin^2⁡2x+2=2 ⇒1-2sin^2⁡2x=0 ⇒sin^2⁡2x=1/2 ⇒2x=nπ±(π)/4 ⇒x=(nπ)/2±(π)/8 Number of solution = 8

Q21 / 60 · JEE Advanced 2015 (Paper 1) · Mathematics

The number of distinct solutions of the equation $\frac{5}{4} \cos^{2} 2 x + \cos^{4} x + \sin^{4} x + \cos^{6} x + \sin^{6} x = 2$ in the interval $[0, 2 π]$ is

Correct answer: 8

Solution

\frac{5}{4} \cos^{2} 2 x + {(\sin^{2} x + \cos^{2} x)}^{2} + {(\sin^{2} x + \cos^{2} x)}^{3} - 5 \sin^{2} x . \cos^{2} x = 2

\Rightarrow \frac{5}{4} (1 - \sin^{2} 2 x) - \frac{5}{4} \sin^{2} 2 x + 2 = 2

\Rightarrow 1 - 2 \sin^{2} 2 x = 0

\Rightarrow \sin^{2} 2 x = \frac{1}{2}

\Rightarrow 2 x = n π \pm \frac{π}{4}

\Rightarrow x = \frac{n π}{2} \pm \frac{π}{8}

Number of solution = 8

Hence, the correct answer is 8.

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Solution

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The number of distinct solutions of the equation $\frac{5}{4} \cos^{2} 2 x + \cos^{4} x + \sin^{4} x + \cos^{6} x + \sin^{6} x = 2$ in the interval $[0, 2 π]$ is