Solve for all $x$ given that $sin(x) + cos(2x) = 1$, where $x$ is an angle on the unit circle.

To solve for all $x$ given that $\sin(x) + \cos(2x) = 1$:

First, use the double-angle identity for cosine: $\cos(2x) = 2\cos^2(x) – 1$

Substitute this into the equation:

$ \sin(x) + 2\cos^2(x) – 1 = 1 $

Rearrange the equation:

$ \sin(x) + 2\cos^2(x) = 2 $

Since $\sin^2(x) + \cos^2(x) = 1$, we have $\cos^2(x) = 1 – \sin^2(x)$

So:

$ \sin(x) + 2(1 – \sin^2(x)) = 2 $

Which simplifies to:

$ \sin(x) + 2 – 2\sin^2(x) = 2 $

Thus:

$ \sin(x) – 2\sin^2(x) = 0 $

Factor out $\sin(x)$:

$ \sin(x)(1 – 2\sin(x)) = 0 $

So $\sin(x) = 0$ or $\sin(x) = \frac{1}{2}$

Therefore, the solutions are:

$ x = n\pi $

and

$ x = \frac{\pi}{6} + 2n\pi $

or

$ x = \frac{5\pi}{6} + 2n\pi $

where $n$ is any integer.

To solve for $x$ given that $sin(x) + cos(2x) = 1$:

Use the identity: $cos(2x) = 2cos^2(x) – 1$

Substituting:

$ sin(x) + 2cos^2(x) – 1 = 1 $

Rearranging:

$ sin(x) + 2cos^2(x) = 2 $

Using $cos^2(x) = 1 – sin^2(x)$:

$ sin(x) + 2(1 – sin^2(x)) = 2 $

Simplify:

$ sin(x) – 2sin^2(x) = 0 $

Factor:

$ sin(x)(1 – 2sin(x)) = 0 $

So $sin(x) = 0$ or $sin(x) = frac{1}{2}$

Therefore:

$ x = npi $

or

$ x = frac{pi}{6} + 2npi $

or

$ x = frac{5pi}{6} + 2npi $

where $n$ is any integer.

To solve for $x$ given $sin(x) + cos(2x) = 1$