solve sin^2(theta)-cos^2(theta)=0 over interval [0, 2pi]

[Striker]

These also need to be solved on the interval from 0 to 2pi

1) sin^2(theta)-cos^2(theta)=0
sine squared theta minus cosine squared theta equals 0.

2) sin2(theta) + sin(theta) =0
sine two theta plus sine theta equals zero.

Thanks for any help that you can give me on these two problems. They are written out incase the shorthand is hard to understand.

 

[pka]

\(\displaystyle \L\begin{array}{l}
\sin ^2 (x) - \cos ^2 (x) = 0\quad \Leftrightarrow \quad \left| {\sin (x)} \right| = \left| {\cos (x)} \right| \\
x = \frac{\pi }{4},\frac{{3\pi }}{4},\frac{{5\pi }}{4},\frac{{7\pi }}{4} \\
\end{array}\)

 

[soroban]

Hello, Striker!

2) Solve on the interval $\displaystyle (0,\,2\pi):\;\;\sin2\theta \,+\,\sin\theta \:=\:0$

Use the identity: $\displaystyle \:\sin(2x) \;=\;2\cdot\sin(x)\cdot\cos(x)$

We have: $\displaystyle \:2\cdot\sin\theta\cdot\cos\theta\,+\,\sin\theta\;=\;0$

Factor: $\displaystyle \:\sin\theta(2\cdot\cos\theta\,+\,1) \;=\;0$

Can you finish it?

 

[morson]

$\displaystyle sin^2\theta - cos^2\theta = -cos(2\theta)$

If $\displaystyle x \in\ (0, 2\pi)$, $\displaystyle 2x \in\ (0, 4\pi)$