# Determine the x angle if 2sin^2x+sin(-x)-1=0

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### 2 Answers

To determine the angle 2sin^x +sin (-x) -1 = 0.

We know sin (-x) = -sinx.

Therefore the given equation could be written as:

2sin^2x -sinx -1 = 0 .

2s^2 - s -1 = 0 is a quadratic in s, where s = sinx.

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

2s(s+1) -1(s+10 = 0.

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

Equate each factor to zero.

s+1 = 0, or 2s-1 = 0.

s+1 = 0 gives s = -1 , sinx = -1, Or x = pi. Or x = (2n+1)pi.

2s-1 = - gives s = 1/2. Or sinx = 1/2. Or = npi +(-1)^n * pi/6.

For the beginning, we remark that one term has as argument the opposite variable, -x. Since the sine function is odd, we'll write the term:

sin(-x) = - sin x

We'll re-write the given expression:

2 (sin x)^2 - sin x - 1 = 0

We'll substitute sin x = t.

We'll re-write the equation using the new variable t:

2t^2 - t - 1 = 0

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

t1 = [1 + sqrt(1 + 8)]/4

t1 = (1+3)/4

t1 = 1

t2 = (1-3)/4

t2 = -1/2

We'll put sin x = t1.

sin x = 1

x = (-1)^k*arc sin 1 + k*pi

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

Now, we'll put sin x = t2.

sin x = -1/2

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

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

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

**The solutions of the equation are: {(-1)^k*(pi/2) + k*pi}U{(-1)^(k+1)*(pi/6) + k*pi}.**