sin(x)-cos(x)=0 find six solutions between 0 to 2pi

Question

Find the solutions of the equation that are in the interval [0, 2π). (Enter solutions from smallest to largest. If there are any unused answer boxes, enter NONE in the last boxes.If the equation has no solution, enter NONE for each answer.)

Answer 1

Change to Sin(x) = Cos(x). Since Sine is y/r and Cos is x/r you know that you're looking on the unit circle where x=r, which occurs at pi/4 and 5pi/4.

Alternatively, square both sides

(sinx)^2 = (cosx)^2

change (cosx)^2 = 1 - (sinx)^2

to get:

(sinx)^2 = 1 - (sinx)^2

rearrange

(sinx)^2 = 1/2

sinx = +- root(2)/2

This occurs at pi/4, 3pi/4, 5pi/4, 7pi/4, however we have too many solutions, because both sin and cos must have same sign( extrenious solutions created by squaring and square rooting), so only at pi/4 and 5pi/4

 

Alternatively, multiply both sides by the conjugate (sinx + cosx).

(sinx)^2 - (cosx)^2 = 0

Use identity (sinx)^2 - (cosx)^2 = cos2x

cos2x = 0

2x = pi/2 and 3pi/2

divide all by 2

x = pi/4 and 3pi/4

Comments

Sorry, made a mistake, we have to double the domain to 0<= 2x <= 4pi
2x = pi/2, 3pi/2, 5pi/2, 7pi/2
Divide by two
x = pi/4, 3pi/4, 5pi/4, 7pi/4
Get rid of the extrenous solutions from (cosx+sinx)
x = pi/4 , 5pi/4