Posted by **Isaac** on Monday, December 5, 2011 at 12:40pm.

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

sin(x/2)+cos(x)=0

- trig -
**Reiny**, Monday, December 5, 2011 at 1:17pm
recall the half-angel conversions

cos 2A = 1 - 2sin^2 A

so

cos x = = 1 - 2sin^2 (x/2)

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

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

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

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

x/2 = 7π/6 or x/2 = 11π/6 or x/2 = π/2

x = 7π/3 or x = 11π/3 or x = π (420° , 660· , 180° )

so for the given domain

x = π

## Answer This Question

## Related Questions

- Math - Can I please get some help on these questions: 1. How many solutions does...
- trig - I need to find all solutions of the given equations for the indicated ...
- trig - 2sin(x)cos(x)+cos(x)=0 I'm looking for exact value solutions in [0, 3&#...
- Math - 1. On the interval [0, 2pi] what are the solutions to the equation ...
- Math Trig - Find all solutions on the interval [0,2pi] for the following: 2sin^2...
- Math Help - Hello! Can someone please check and see if I did this right? Thanks...
- trig - Solve cos x-1 = sin^2 x Find all solutions on the interval [0,2pi) a. x=...
- trig - Find all solutions of cos (x) + 1/2 sec (x) = -3/2 in the interval (2pi, ...
- trig equations - Find all solutions (if they exist)for the given equation on the...
- Trig - If cosx = 10/19 an pi < x < 2pi, find the exact value of cos x/2 ...

More Related Questions