How do you solve: sin2θ + cosθ = 0 between 0 and 2pi?

Apr 27, 2018

$\theta = \frac{\pi}{2} , \frac{3 \pi}{2} , \frac{7 \pi}{6} , \frac{11 \pi}{6}$

Explanation:

We're going to want to get rid of the double angle, $\sin 2 \theta .$

Recall the identity $\sin 2 \theta = 2 \sin \theta \cos \theta$ and apply it:

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

Note that all of these terms involve $\cos \theta ,$ so we can factor it out:

$\cos \theta \left(2 \sin \theta + 1\right) = 0$

We now solve the following equations:
$\cos \theta = 0$
$2 \sin \theta + 1 = 0$

Each coordinate pair $\left(x , y\right)$ on this unit circle represents $\left(\cos \theta , \sin \theta\right)$ at an angle $\theta .$

For $\cos \theta = 0 ,$ we see from the unit circle that this holds true for $\theta = \frac{\pi}{2} , \frac{3 \pi}{2}$ over the interval $\left[0 , 2 \pi\right) .$

$2 \sin \theta + 1 = 0$
$2 \sin \theta = - 1$
$\sin \theta = - \frac{1}{2}$

We see this holds true for $\theta = \frac{7 \pi}{6} , \frac{11 \pi}{6}$ over $\left[0 , 2 \pi\right) .$

Apr 27, 2018

I tried this:

Explanation:

We can write:

$\sin 2 \theta = 2 \sin \theta \cos \theta$

so that our equation becomes:

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

collect:

$\cos \theta \left(2 \sin \theta + 1\right) = 0$

so, this equation is satisfied either when:

$\cos \theta = 0$
i.e. when $\theta = \frac{\pi}{2} \mathmr{and} \theta = \frac{3}{2} \pi$

or when:
$2 \sin \theta + 1 = 0$
$\sin \theta = - \frac{1}{2}$
i.e. when $\theta = \frac{7}{6} \pi \mathmr{and} \theta = \frac{11}{6} \pi$

Apr 27, 2018

$\theta \in \left\{\frac{\pi}{2} , \frac{7 \pi}{6} , \frac{11 \pi}{6} , \frac{3 \pi}{2}\right\}$

Explanation:

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

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

$\cos \theta \left(2 \sin \theta + 1\right) = 0$

$\cos \theta = 0$ or $2 \sin \theta + 1 = 0$

$\cos \theta = 0$ or $\sin \theta = - \frac{1}{2}$

You knew a 30/60/90 or 45/45/90 triangle was going to show up sooner or later.

$\theta = \frac{\pi}{2} + \pi k \quad \mathmr{and} \quad \theta = - \frac{\pi}{6} + 2 \pi k \quad \mathmr{and} \quad \theta = - \frac{5 \pi}{6} + 2 \pi k$

The last two came from two supplementary angles ($- {30}^{\circ}$ and $- {150}^{\circ}$) whose sines are $- \frac{1}{2.}$

Let's enumerate $0 \setminus \le \theta \setminus \le 2 \pi$.

From $\setminus \frac{\pi}{2} + \setminus \pi k$ we get $\theta = \frac{\pi}{2} , \frac{3 \pi}{2}$

From $- \frac{\pi}{6} + 2 \pi k$ at $k = 1$ we get $\theta = \frac{11 \pi}{6}$

From $\frac{- 5 \pi}{6} + 2 \setminus \pi k$ at $k = 1$ we get $\theta = \frac{7 \pi}{6}$

Check: I'll check a couple, leave the rest for you.

theta={3pi}/2 quad quad sin(3pi)+cos({3pi}/2) = 0+0 quad sqrt

theta={7pi}/6 quad quad sin({7pi}/3)+cos({7pi}/6) =sin(pi/3)-cos(pi/6)=0 quad sqrt

That's the end, but I'll keep writing so feel free to keep reading this optional part..

Getting those $\pi k$ and $2 \pi k$ right is a bit of an art. It helps to go back to first principles, which for me is $\cos x = \cos a$ has solutions $x = \setminus \pm a + 2 \setminus \pi k \quad$ integer $k$.

We rewrite $\cos \theta = 0$ or $\sin \theta = - \frac{1}{2}$ as

$\cos \theta = \cos \left(\frac{\pi}{2}\right)$ or $\cos \left(\frac{\pi}{2} - \theta\right) = \cos \left(\frac{2 \pi}{3}\right)$

Applying our solutions, $\cos \theta = \cos \left(\frac{\pi}{2}\right)$ gives $\theta = \pm \frac{\pi}{2} + 2 \pi k$. That's really two lists, that can be combined giving $\theta = \frac{\pi}{2} + \pi k$. The same set of values is produced for different $k$s.

$\cos \left(\frac{\pi}{2} - \theta\right) = \cos \left(\frac{2 \pi}{3}\right)$ gives

$\frac{\pi}{2} - \theta = \setminus \pm \frac{2 \pi}{3} + 2 \pi k$

$\theta = \frac{\pi}{2} \setminus \pm \frac{2 \pi}{3} + 2 \setminus \pi k$

$\theta = \frac{7 \pi}{6} + 2 \setminus \pi k$ or $\theta = - \frac{\pi}{6} + 2 \pi k$

which is essentially what we got.