4 5 cos 2 2 x + cos 4 x + sin 4 x + cos 6 x + sin 6 x = 2
If x is a real number that satisfy the equation above, find the number of distinct solution(s) of x in the interval [ 0 , 2 π ] .
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cos ( 4 x ) = 0 2 cos 2 ( 2 x ) − 1 cos 2 ( 2 x ) = 2 1 ( 2 cos 2 x − 1 ) 2 = 2 1 cos 4 x − cos 2 x = − 8 1 Let cos 2 x = x . Then: x 2 − x + 8 1 = 0 x = 2 1 − 2 2 1 and x = 2 1 + 2 2 1 First case, cos 2 x = 2 1 − 2 2 1 cos x = 2 1 − 2 2 1 So x = 2 n π ± cos − 1 ( ± 2 1 − 2 2 1 ) So in all, 4 solutions and similarly for the second case 4 solutions.
Cos 4x=0 x=(2n-1)(pi/2) Put n=1 to 8 so 8 solutions
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
Perfect Solution! . used exactly the same approach
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4 5 cos 2 2 x + cos 4 x + sin 4 x + cos 6 x + sin 6 x = 4 5 cos 2 2 x + 1 − 2 sin 2 x cos 2 x + cos 4 x + sin 4 x − sin 2 x cos 2 x = 4 5 cos 2 2 x + 1 − 2 1 sin 2 2 x + 1 − 2 sin 2 x cos 2 x − 4 1 sin 2 2 x = 2 + 4 5 cos 2 2 x − 2 1 sin 2 2 x − 2 1 sin 2 2 x − 4 1 sin 2 2 x = 2 + 4 5 cos 2 2 x − 4 5 sin 2 2 x = 2 + 4 5 cos 4 x = 2 ⇒ cos 4 x = 0
For the domain [ 0 , 2 π ] , the function y = cos 4 x cuts the x axis at 8 points.
Hence the number of distinct solutions = 8