Solve the following equation and state the general solution for all values of x in exact form. Show all steps of your algebraic solution.

Question

𝑠𝑖𝑛^2(𝑥) − √2𝑐𝑜𝑠(𝑥) = 𝑐𝑜𝑠^2(𝑥) + √2𝑐𝑜𝑠(𝑥) + 2

I find this question to be really difficult. Could anyone help me?

Answer 1

since sin^2x = 1-cos^2x, let u=cosx and
make things easier to read and write it as
1 - u^2 - √2 u = u^2 + √2 u + 2
2u^2 + 2√2 u + 1 = 0
(√2 u + 1)^2 = 0
u = -1/√2
so now we have
cosx = -1/√2
x = π ± π/4 + 2kπ

Answer 2

change the sin^2 x to 1 - cos^2 x and you have only cos x in your equation:
1 - cos^2 x - √2cosx = cos^2 x + √2cosx + 2
2cos^2 x + 2√2cosx + 1 = 0
This factors to
(√2 cosx + 1)^2 = 0
cosx = -1/√2
we know the cosine is negative in quads II and III
x = 135° or x = 225°
or in radians, x = 3π/4, x = 5π/4

The period of cosx is 2π
so we have:
x = 3π/4 + 2kπ , 5π/4 + 2kπ, where k is an integer.