Solve $(\sin (\theta)+\cos(\theta))^2= \frac{3}{2}$

I came across this problem:

Solve for $$\theta$$ (between value $$0$$ and $$2\pi$$) where $$(\sin (\theta)+\cos(\theta))^2= \frac{3}{2}$$

I started by expanding the left side to get $$\sin^2(\theta)+\cos^2{\theta}+\sin(\theta)\cos(\theta) = \frac{3}{2}$$ then changed sin&cos squared sum into $$1$$ and used double angle formula for sin to get $$\sin(2\theta)=1$$. Then I continued to get the solution $$\theta=\pi/4 +2\pi n$$ but the answer is incorrect.

Can I get help on understanding my mistake here?

• The first expansion of the left side is wrong. Commented Mar 10 at 12:24
• @jjagmath Thank you. I really thought I was missing something important. Commented Mar 10 at 13:36

Your thought process is correct overall but you've made a mistake while expanding the left side. More precisely, you should've done

$$(\sin(\theta) + \cos(\theta))^2 = \sin^2(\theta) + 2 \sin(\theta)\cos(\theta) + \cos^2(\theta) = 1 + 2\sin(\theta)\cos(\theta) = 1 + \sin(2\theta).$$

Hence, it follows that

$$(\sin(\theta) + \cos(\theta))^2 = \frac{3}{2} \Longleftrightarrow \sin(2\theta) = \frac{1}{2} .$$

Solving the latter equation, you obtain

$$2\theta = \frac{\pi}{6} + 2\pi n \quad \text{ or } \quad 2\theta = \frac{5\pi}{6} + 2\pi n,$$

which is equivalent to

$$\theta = \frac{\pi}{12} + \pi n \quad \text{ or } \quad \theta = \frac{5\pi}{12} + \pi n.$$

Since you're only interested in values of $$\theta$$ such that $$\theta \in [0,2\pi]$$, your solutions are

$$\Big\{ \frac{\pi}{12}, \frac{5\pi}{12}, \frac{13\pi}{12},\frac{17\pi}{12} \Big\}.$$

• You forgot about $\theta = \frac{13 \pi}{12}$. Commented Mar 10 at 12:19
• Should be fine now @BenSteffan
– xyz
Commented Mar 10 at 12:30