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cos2x can be written as 1 - 2 sin2x so the equation becomes
1 - 2 sin2x +a sin x + 7 - 2a = 0
2 sin2x -a sin x - 8 + 2a =0
it is quadratic in sin x
so by using quadratic formula
we get
sin x = a +-√a2-4 * 2(2a - 8)/4\
on solving we get
sin x = (a - 4 )/2
to find the possible values of a we will use the following inequation
as we know that value of sin x lies between -1 to 1
-1 <=(a - 4 )/2<=1
-2 <=(a - 4 )<=2
2 <= a <=6
so for this range the solution of the trigonometric equation exists
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