Find all principal solutions

Geometry Level 4

5 4 cos ⁡ 2 2 x + cos ⁡ 4 x + sin ⁡ 4 x + cos ⁡ 6 x + sin ⁡ 6 x = 2 \large\dfrac{5}{4}\cos^{2}2x+\cos^{4}x+\sin^{4}x+\cos^{6}x+\sin^{6}x=2

Find the number of solutions of the above equation lying in the interval [ 0 , 2 π ] \left[0,2\pi\right] .


The answer is 8.

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Tapas Mazumdar by Tapas Mazumdar , 20, India

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1 solution

LHS = 5 4 cos ⁡ 2 2 x + 1 − 2 sin ⁡ 2 x cos ⁡ 2 x + 1 − 3 sin ⁡ 2 x cos ⁡ 2 x = 5 4 cos ⁡ 2 2 x + 2 − 5 4 sin ⁡ 2 2 x = 5 4 cos ⁡ 4 x + 2 \displaystyle \text{LHS}= \frac{5}{4}\cos^2 2x+1-2\sin^2 x \cos^2 x + 1-3\sin^2 x \cos^2 x = \frac{5}{4}\cos^2 2x +2-\frac{5}{4}\sin^2 2x = \frac{5}{4}\cos 4x +2

So, 5 4 cos ⁡ 4 x + 2 = 2 ⟹ cos ⁡ 4 x = 0 \displaystyle \frac{5}{4}\cos 4x +2=2 \implies \cos 4x=0 so , x = ( 2 n + 1 ) π 8 \displaystyle x=\frac{(2n+1)\pi}{8} which says there are 8 values of n n for which it lies in the interval [ 0 , 2 π ] [0,2\pi] so number of solutions is 8 \boxed{8}

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JEE 2015 Integer type?

Prakhar Bindal - 4 years, 9 months ago

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