JEE-Advanced 2015 (1/40)

Geometry Level 4

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

If x x is a real number that satisfy the equation above, find the number of distinct solution(s) of x x in the interval [ 0 , 2 π ] [0,2\pi] .


The answer is 8.

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Sandeep Bhardwaj by Sandeep Bhardwaj , 28, India

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1 solution

Rohit Ner
Jun 16, 2015

5 4 cos ⁡ 2 2 x + cos ⁡ 4 x + sin ⁡ 4 x + cos ⁡ 6 x + sin ⁡ 6 x = 5 4 cos ⁡ 2 2 x + 1 − 2 sin ⁡ 2 x cos ⁡ 2 x + cos ⁡ 4 x + sin ⁡ 4 x − sin ⁡ 2 x cos ⁡ 2 x = 5 4 cos ⁡ 2 2 x + 1 − 1 2 sin ⁡ 2 2 x + 1 − 2 sin ⁡ 2 x cos ⁡ 2 x − 1 4 sin ⁡ 2 2 x = 2 + 5 4 cos ⁡ 2 2 x − 1 2 sin ⁡ 2 2 x − 1 2 sin ⁡ 2 2 x − 1 4 sin ⁡ 2 2 x = 2 + 5 4 cos ⁡ 2 2 x − 5 4 sin ⁡ 2 2 x = 2 + 5 4 cos ⁡ 4 x = 2 ⇒ cos ⁡ 4 x = 0 \frac{5}{4} \cos^2 2x+\cos^4 x+\sin^4 x+\cos^6 x+\sin^6 x \\ =\frac { 5 }{ 4 } \cos ^{ 2 }{ 2x } +1-2\sin ^{ 2 }{ x } \cos ^{ 2 }{ x } +\cos ^{ 4 }{ x } +\sin ^{ 4 }{ x } -\sin ^{ 2 }{ x } \cos ^{ 2 }{ x } \\ =\frac { 5 }{ 4 } \cos ^{ 2 }{ 2x } +1-\frac { 1 }{ 2 } \sin ^{ 2 }{ 2x } +1-2\sin ^{ 2 }{ x } \cos ^{ 2 }{ x } -\frac { 1 }{ 4 } \sin ^{ 2 }{ 2x } \\ =2+\frac { 5 }{ 4 } \cos ^{ 2 }{ 2x } -\frac { 1 }{ 2 } \sin ^{ 2 }{ 2x } -\frac { 1 }{ 2 } \sin ^{ 2 }{ 2x } -\frac { 1 }{ 4 } \sin ^{ 2 }{ 2x } \\ =2+\frac { 5 }{ 4 } \cos ^{ 2 }{ 2x } -\frac { 5 }{ 4 } \sin ^{ 2 }{ 2x } \\ =2+\frac { 5 }{ 4 } \cos { 4x } =2 \\ \Rightarrow \cos { 4x } =0

For the domain [ 0 , 2 π ] \left[ 0,2\pi \right] , the function y = cos ⁡ 4 x y=\cos { 4x } cuts the x x axis at 8 points.

Hence the number of distinct solutions = 8 \Huge\color{#3D99F6}{=\boxed{8}}

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Can you solve this without graphing?

cos ⁡ ( 4 x ) = 0 \cos(4x)=0 2 cos ⁡ 2 ( 2 x ) − 1 2\cos^2(2x)-1 cos ⁡ 2 ( 2 x ) = 1 2 \cos^2(2x)=\frac12 ( 2 cos ⁡ 2 x − 1 ) 2 = 1 2 (2\cos^2x-1)^2=\frac12 cos ⁡ 4 x − cos ⁡ 2 x = − 1 8 \cos^4x-\cos^2x=-\frac18 Let cos ⁡ 2 x = x \cos^2x=x . Then: x 2 − x + 1 8 = 0 x^2-x+\frac18=0 x = 1 2 − 1 2 2 x=\frac12-\frac1{2\sqrt{2}} and x = 1 2 + 1 2 2 x=\frac12+\frac1{2\sqrt{2}} First case, cos ⁡ 2 x = 1 2 − 1 2 2 \cos^2x=\frac12-\frac1{2\sqrt{2}} cos ⁡ x = 1 2 − 1 2 2 \cos x=\sqrt{\frac12-\frac1{2\sqrt{2}}} So x = 2 n π ± cos ⁡ − 1 ( ± 1 2 − 1 2 2 ) x=2n\pi\pm\cos^{-1}(\pm\sqrt{\frac12-\frac1{2\sqrt{2}}}) So in all, 4 4 solutions and similarly for the second case 4 4 solutions.

Aditya Agarwal - 5 years, 9 months ago

Cos 4x=0 x=(2n-1)(pi/2) Put n=1 to 8 so 8 solutions

Aqid Khatkhatay - 4 years, 5 months ago

Substitute \cos{4x}=1- 2\sin^2 {2x} and you will get

\sin^2 {2x} =\frac{1}{10}

Where 2x is in interval [0 , 4\pi] which we know will happen 8 times. Simple

Anurag Pandey - 4 years, 2 months ago

Perfect Solution! . used exactly the same approach

Prakhar Bindal - 5 years, 12 months ago

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