Question

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

sinx=-cos²x-1
Write your answer in radians in terms of T.

Answer 1

x = 3π/2

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)