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Question

Find number of value of xϵ[0,π] satisfying the relation cos 3x + sin 2x - sin 4x = 0


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Solution

We have to simplify the given expression to solve it. By looking at the expression, we see that we can

apply transformation formula for sin 2x — sin 4x.

cos 3x + sin 2x - sin 4x = 0

cos 3x + 2 sin (-x) cos 3x = 0

cos 3x (1 - 2 sin x) = 0

cos 3x = 0 or (1-2 sin x) = 0
cos 3x = 0 or sin x = 12
3x = (2n + 1) π2 or x=nπ + (1)n π6
x = (2n + 1) π6 or x=nπ+ (1)n π6
By using these two expressions we see that the value of x which lie in [0,π] in are π6 , π3 , π3 (we get these
values by substituting n = 1, 2, 3 .... in the expressions we got for x)
So we get three values in [0,π]


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