I have two questions how do you solve cos x=sqrt2-cosx on the interval of [0,2pi) and how do you verify sin^2(theta)/1-cos(theta)=1+cos(theta) without negating it to equal 1?

Answer 1

1.
cosx = √2 - cosx
2cosx = √2
cosx = √2/2 which is the same as 1/√2
if you are familiar with the 1,1,√2 right-angled triangle you should know that x = 45° or π/4 radians

(you can use your calculator to verify this)

but the cosine is positve in quadrants I and IV
so x = π/4 or x = 2π - π/4 = 7π/4

2.
For this one you should know that
sin^2 Ø + cos^2 Ø = 1 or
sin^2 Ø = 1 - cos^2 Ø

LS = sin^2Ø/(1- cosØ)
= (1 - cos^2 Ø)/(1-cosØ)
= (1+cosØ)(1-cos‚/(1-cosØ)
= 1 + cosØ
= RS

Answer 2

Thank you so much!