# Evaluating solutions for a given trigonometric equation over a specified interval

This question is similar in form to this one:

Finding all Trigonometric Solutions of an Equation within a Given Interval

However, I want to verify my method of solving, as it would appear I have made some logical error in my processes.

My method goes as follows:

Solve for $\theta$ over the interval $[0, 2\pi)$

$\sqrt{2}\sin{\theta} = \sqrt{\sin{\theta}+1}$

$(\sqrt{2}\sin{\theta})^2 = (\sqrt{\sin{\theta}+1})^2$

$2\sin^2{\theta} = \sin{\theta} + 1$

$\sin^2{\theta} = \dfrac{\sin{\theta} + 1}{2}$

$\dfrac{1-\cos{2\theta}}{2} = \dfrac{\sin{\theta} + 1}{2}$

$1-\cos{2\theta} = \sin{\theta} + 1$

$-\cos{2\theta} = \sin{\theta}$

$0 = \sin{\theta} + \cos{2\theta}$

Then solve for $\theta$.

$\sin{\dfrac{1\pi}{2}} = 1$

$\cos{\dfrac{2\pi}{2}} = \cos{\pi} = -1$

Therefore $\theta = \dfrac{\pi}{2}$

I was told my method was incorrect. I don't see where.

## 1 Answer

I see two problems with your solution.

First, how did you get from the equation $0 = \sin{\theta} + \cos{2\theta}$ to concluding that $\theta = \dfrac{\pi}{2}$? You merely showed that it is one solution: you did not show that is the only solution. And in fact it is not the only solution. Other solutions to $0 = \sin{\theta} + \cos{2\theta}$ are $\frac{7\pi}6$, $\frac{11\pi}6$ and any values of $\theta$ for which

$$\sin\theta=-\frac 12$$

In fact, you could have gotten those other solutions from the earlier equation $2\sin^2{\theta} = \sin{\theta} + 1$ by treating that as a quadratic equation in $\sin\theta$ and solving. The solutions are $\sin\theta\in\{-\frac 12,1\}$.

The second problem is that you added an extraneous solution when you squared both sides of your original equation. That extraneous solution is in fact $\sin\theta=-\frac 12$. The fact that you did not see that extraneous solution seems to be a stroke of luck, not as a result of you seeing and then excluding that false answer.

In summary, a better way to solve your original equation would have been to solve the quadratic equation $2\sin^2{\theta} = \sin{\theta} + 1$ and testing those solutions in your original equation. You would have ended up with the same answer you got, but in a proper way.

• $-\dfrac{\pi}{6}$ is not within the interval $[0,2π)$. – user146046 Feb 7 '15 at 2:51
• I probably should have made it more clear, but I meant the multivalued arcsine: any value of $\theta$ for which $\sin\theta=-1/2$. In your interval that would be $7\pi/6$ and $11\pi/6$. I'll edit my answer for greater clarity. – Rory Daulton Feb 7 '15 at 11:08