# General solutions of Trigonometric functions

Find the general solution of:

$$\sin^3 \theta - \sin \theta = 0$$

Working out: (Factorise out)

$$\sin \theta (\sin^2 \theta - 1) = 0$$

Solve for $\sin \theta$ and $\sin^2 \theta - 1$:

For $\sin \theta = 0$

$$\sin \theta = 0$$

$$\therefore \theta = 0$$

$$\therefore \theta = n \pi$$

For $\sin^2 \theta - 1 = 0$:

$$\sin^2 \theta - 1 = 0$$

$$\sin^2 \theta = 1$$

$$\sin \theta = \pm1$$

$$\theta = - \pi/2, \pi/2$$

$$\therefore \theta = n\pi + (-1)^n(\pm\pi/2)$$

So i get the two solutions as:

$$\theta = n\pi\text{ and }\theta = n\pi + (-1)^n(\pm\pi/2)$$

However a complication occurs because wolframalpha says otherwise, help would be greatly appreciated.

Wolframalpha

• Hint: $sin^2\theta - 1 = - cos^2\theta$ – Lost Dec 10 '13 at 4:30
• I would write $n\pi +\frac{\pi}{2}$ for the $\sin^2\theta=1$ case. Or maybe $2n\pi \pm \frac{\pi}{2}$. You are entioning everybody twice. Which is not wrong, but it is not optimal. – André Nicolas Dec 10 '13 at 4:33
• @AndréNicolas Can someone explain to me why my one is wrong though? – MATHSUSER Dec 10 '13 at 4:36
• @Lost Why would I make it equal to $-\cos^2 \theta$?, why not just use what's provided? – MATHSUSER Dec 10 '13 at 4:37
• Take $n=0$, and the $+$. We get $\frac{\pi}{2}$. Take $n=1$, and the $+$. We get $\frac{\pi}{2}$. Right, but stuttery. You should also not talk about $2$ solutions, it is $2$ (or $3$, depends how one counts) families of solutions. – André Nicolas Dec 10 '13 at 4:38

## 2 Answers

$\displaystyle\cos^2\theta=0$ is definitely the easiest as it implies and is impled by $\displaystyle\cos\theta=0\implies \theta$ is odd multiple of $\displaystyle \frac\pi2$

Otherwise using Double-Angle Formulas,

$\displaystyle\sin^2\theta=\sin^2\alpha\iff \cos^2\theta=\cos^2\alpha$ $\displaystyle\iff\tan^2\theta=\tan^2\alpha$ $\displaystyle\iff\cos2\theta=\cos2\alpha$

$\displaystyle\implies 2\theta=2n\pi\pm2\alpha$ where $n$ is any integer

When you write the solution for $$\ (\sin^2 \theta - 1) = 0$$ your answer is correct but you did not notice that what you wrote is exactly an odd multiple of $\displaystyle \frac\pi2$ as mentioned by lab bhattacharjee.