Question

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

sin 2x + cos x = 0
X =
X =
T
X =
2
x = 120
T
2
+
T
2
2nx
3
+ na
+ na
X
X
X
X
(smallest value)
(largest value)

Answer 1

We can use trigonometric identities to rewrite the equation as:

2sin x cos x + cos x = 0

Factoring out cos x, we get:

cos x (2sin x + 1) = 0

This equation has two solutions in the interval [0, 2π):

cos x = 0 when x = π/2 and x = 3π/2

2sin x + 1 = 0 when sin x = -1/2, which occurs at x = 7π/6 and x = 11π/6.

However, we need to find the solutions in the interval [0, 2x). Since the period of both sin x and cos x is 2π, we can add any multiple of 2π to the solutions we found to get all possible solutions in the interval [0, 2x).

For the solutions x = π/2 and x = 3π/2, we have:

π/2, 5π/2 (adding 2π)

3π/2, 7π/2 (adding 2π)

For the solutions x = 7π/6 and x = 11π/6, we have:

7π/6, 19π/6 (adding 2π)

11π/6, 23π/6 (adding 2π)

The smallest solution in the interval [0, 2x) is π/2, and the largest solution is 23π/6.

Therefore, the solutions are:

x = π/2, 3π/2, 7π/6, 11π/6, 19π/6, 23π/6.