Solve [0, 2pi)
3 Cos(2x) = 5 Sin(x) - 1

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.

in problems involving proportions, we can use cross products to test whether two ratios are equal and form a proportion. to find the cross products of a proportion, we multiply the outer terms, called the extremes, and the middle terms, called the means.

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