Question

Solve the equation of the interval (0, 2pi)

cosx=sinx

I squared both sides to get :cos²x=sin²x

Then using tri indentites I came up with

cos²x=1-cos²x

Ended up with 2cos²x=1

Would the answer be cos²x=1/2???

Answer 1

so far so good, divide both sides by 2,
now take √ of both sides,

cos x = ± 1/√2
so x = 45, 135, 225 and 315º
the problem is that you squared both sides, so each answer must be verified.
only 60º and 300º work

so x = 45 or x= 225, or in radians
x = pi/4 or x = 5pi/4

another ways that would have avoided the above problems, and the way I would have done it, is to divide both sides by cosx
sinx/cosx = 1
tanx = 1
now x = 45º or x = 180+45 = 225º
or x = pi/4 or x = 5pi/4

Answer 2

the 6th line down which was
<< only 60º and 300º work >>

should have said :

only 45º and 225º work

Answer 3

So my best way to solve

(Tanx +1)(cosx +1)=0

Would be to divide by either cos or tan..instead of squaring and using an identity?

Answer 4

You don't have to do any more before proceeding.
since you have a product equal to zero, simply set each of its factors to zero, so ...
tanx + 1 = 0 or cosx + 1 = 0

1. tanx = -1 , so x must be in quadrants II or IV and
x = 180-45 or x = 360-45
x = 135 or 315

2. cosx = -1
x = 180

so x = 135 , 180 or 315 which in radians is
x = 3pi/4 , pi , 7pi/4