# Solve trigonometric equation 2sin^2x - 3sinx = - 1.

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To solve 2sin^2x-3sinx= -1.

This is a quadratic equation in sinx.

So we put sinx = s in the given equation and solve for s first. And then from the solution we determine x.

2s^2 -3s = -1.

We add 1 to both sides:

2s^2-3s+1 = 0

We factorise the left:

2s^2 -2s-s+1 = 0

2s(s-1) -1(s-1) = 0

(s-1)(2s-1) = 0

s-1 = 0 or 2s-1 = 0

s-1 = 0 gives s = 1. So sinx = 1, or x = pi/2. Or x = 2npi+pi/2

2s-1 = 0 gives s = 1/2 . x = pi/6 or 5pi/6. Or x = np +(-1)^n*pi/6.

The first step is to move all terms to the left side:

2(sin x)^2- 3sinx + 1 = 0

Now, we'll use substitution technique to solve the equation.

We'll note sin x = t and we'll re-write the equation in t:

2t^2 - 3t + 1 = 0

Since it is a quadratic, we'll apply the quadratic formula:

t1 = {-(-3) + sqrt[(-3)^2 - 4*2*1]}/2*2

t1 = [3+sqrt(9-8)]/4

t1 = (3+1)/4

t1 = 1

t2 = (3-1)/4

t2 = 1/2

Now, we'll put sin x = t1.

sin x = 1

Since it is an elementary equation, we'll apply the formula:

sin x = a

x = (-1)^k* arcsin a + 2k*pi

In our case, a = 1:

x = (-1)^k* arcsin 1 + 2k*pi

x = (-1)^k*(pi/2) + 2k*pi

x = pi/2

Now, we'll put sin x = t2

sin x = 1/2

x = (-1)^k* arcsin 1/2 + 2k*pi

x = (-1)^k* (pi/6) + 2k*pi

x = pi/6

x = pi - pi/6

x = 5pi/6

**The solutions of the equation are:{ pi/6 ; pi/2 ; 5pi/6 }.**