How Do You Solve Trigonometric Equations Involving Sine and Cosine?

[zeion]

Homework Statement

Solve the following equations of inequalities in the interval [0, 2pi)

cosx + 1 = sinx

Homework Equations

The Attempt at a Solution

cos²x + 2cosx + 1 = sin²x
cos²x + 2cosx + 1 = 1 - cos²x
2cos²x + 2cosx = 0
(2cosx)(cosx + 1) = 0
2cosx = 0 or cosx + 1 = 0
2cosx = 0
cosx = 0
x = pi/2, 3pi/2

cosx + 1 = 0
cosx = -1
x = pi

Did I miss anything?

 

[mrkuo]

You forgot to check your answers. When squaring an equation, you create the possibility of extraneous solutions appearing. In the case of your problem, 3pi/2 does not work in the original equation and should be left out of the solution set.

 

[Chewy0087]

yeh you can't be too careful squaring both sides as you risk getting extra solutions, perhaps a 'cleaner' way to do it is simply to rearrange giving;

sinx - cosx = 1

then proceed to right it in the form Rsin(A-B)= 1

but your method works, like mrko said just remember to check