Given :
Solve [0, 2π)
3cos(2x) = 5sin(x) - 1
To Find :
The value of x in range [0,2π).
Solution :
We know, cos(2x) = 1 - sin²x .
3( 1 - 2sin²x ) = 5sinx - 1
3 - 6sin²x = 5sinx - 1
6sin²x + 5sinx - 4 = 0
6sin²x + 8sinx - 3sinx - 4 = 0
2sinx( 3sinx + 4 ) - 1( 3sinx + 4) = 0
sin x = 1/2 and sin x = -4/3 .
Since, sin x ≤ 1.
So, sin x = 1/2 .
x = π/6 .
Therefore, value of x is π/6.
Hence, this is the required solution.
