Question

Find all solutions of the equation in the interval [0, 2).

Answer 1

Answer:

  x = 3π/2

Step-by-step explanation:

The given equation can be rewritten as a quadratic in sin(x), then solved by factoring.

Rewrite

Using the identity ...

  cos²x = 1 -sin²x

the equation can be rewritten as ...

  sin(x) = -(1 -sin²(x)) -1 . . . . . .substitute for cos²x

Solution

The new equation can be solved by considering it a quadratic in sin(x).

  0 = sin²(x) -sin(x) -2 . . . . . . subtract sin(x)

  (sin(x) -2)(sin(x) +1) = 0 . . . . factor quadratic

The solutions to this are ...

  sin(x) -2 = 0   ⇒   sin(x) = 2 . . . .  no solutions

  sin(x) +1 = 0   ⇒   sin(x) = -1   ⇒   x = 3π/2 (one solution)