Hi Kayla,

Let's look at the first equation.

sin2x = sinx

2sinxcosx = sinx (trig identity)

2sinxcosx - sinx = 0 (subtraction)

sinx(2cosx - 1) = 0 (factor)

sinx = 0 or 2cosx -1 = 0 (zero product property)

x=sin^{-1}0 or cosx = 1/2

x = 0, π, 2π..., which we can write kπ where k is any integer.

For x = cos^{-1}1/2 = π/3, -π/3 ... (to list all solutions: π/3 + 2kπ, k any integer or -π/3+ 2kπ, k any integer)

There are several different ways to represent the set of all solutions, but this gives you the idea.

For 2)

cos^{2}x + cos2x = -1

cos^{2}x + (cos^{2}x -1) = -1 (trig identity; since cos already appears in the equation choose to use this flavor of cos2x)

2cos^{2}x = 0 (add 1)

cos^{2}x = 0 (division)

cosx = 0 (square root)

x= cos^{-1}0 = π/2 and -π/2, but there are infinitely many more solutions. One way to write that is:

x = π/2 + kπ, k any integer