Solve for x!!!!!!!!!!!!!!!!!!!!!!!!

Question

2cos^2x - 2sin^2x -2sinx =0 I'd like an explanation as to how you solve for x please.

Answer 1

2cos²x - 2sin²x - 2sinx = 0

Divide each side by 2.

cos²x - sin²x - sinx = 0

Pythagorean Identities: sin²θ + cos²θ = 1 ⇒ cos²θ = 1 - sin²θ

1 - sin²x - sin²x - sinx = 0

1 - 2sin²x - sinx = 0

Multiply each side by negative one.

2sin²x + sinx - 1 = 0

Let sinx = t

2t² + t - 1 = 0

Now solve the equation using the factor method.

2t² + 2t - t - 1 = 0

2t(t + 1) -1(t + 1) = 0

(2t - 1)(t + 1) = 0

2t - 1 = 0 or t +1 = 0

2t - 1 = 0 ⇒ t = 1/2

t + 1 = 0 ⇒ t = -1

But t = sinx

If t = 1/2 ⇒ sinx = 1/2 ⇒ x = arc sin(1/2) ⇒x = 30°= π/6

If t = -1 ⇒ sinx = -1 ⇒ x = arc sin(-1) ⇒ x = 270°= 3π/2

Therefore x = π/6 or 3π/2.

Comments

The solutions are x = 2nπ + 3π/2, x = 2nπ + π/6, and x = 2nπ + 5π/6, where n∈ Z.

Answer 2

The trigonometric equation is 2cos²x - 2sin²x - 2sin x = 0.

Divide each side by 2.

cos²x - sin²x - sin x = 0

Pythagorean identity : cos²x + sin²x = 1.

(1 - sin²x) - sin²x - sin x = 0

1 - 2sin²x - sin x = 0

2sin²x + sin x - 1 = 0.

By factor by grouping.

2sin²x + 2sin x - sin x - 1 = 0

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

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

Apply zero product property.

sin x +1 = 0     or       2sin x - 1 = 0

sin x = - 1        or       sin x = 1/2

• sin (x) = - 1.

The function sin (x) has a period of 2π, first find all solutions in the interval [0, 2π).

The function sin (x) is negative in third and fourth quadrant.

sin(x) = - 1.

In third Quadrant, π ≤ x ≤ 3π/2.

- 1 = sin (- π/2) = sin (π + π/2) = sin (3π/2).

In fourth Quadrant, 3π/2 ≤ x ≤ 2π.

- 1 = sin (- π/2) = sin (2π - π/2) = sin (3π/2).

Finally, add multiples of to each of these solutions to get the general form.

x = 2nπ + 3π/2, where n∈ Z.

• sin (x) = 1/2.

The function sin (x) has a period of 2π, first find all solutions in the interval [0, 2π) .

The function sin (x) is positive in first and second quadrant.

sin(x) = 1/2.

In first Quadrant, 0 ≤ x ≤ π/2.

1/2 = sin (π/6) = sin (0 + π/6) = sin (π/6).

In second Quadrant, π/2 ≤ x ≤ π.

1/2 = sin (π/6) = sin (π - π/6) = sin (5π/6).

Finally, add multiples of to each of these solutions to get the general form.

x = 2nπ + π/6 and x = 2nπ + 5π/6 where n∈ Z.

 

So, the general solutions are x = 2nπ + 3π/2, x = 2nπ + π/6, and x = 2nπ + 5π/6, where n∈ Z.