Solve `cos 2x=|sin x|, x in (-pi/2, pi)`.
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Correct Answer – `x=-pi/6, pi/6, (5pi)/6`
If `sin x gt 0`, then we have
`2 sin^(2)x+sin x -1 =0`
`rArr sin x=1/2`,
`:. x=pi/6, (5pi)/6`
If `sin x lt 0` then we have
`2sin^(2) x- sin x-1=0`
`rArr sinx=-1/2`
`:. x=-pi/6`