Solve $\sin(x)=-\frac{1}{2}$

Question

I have to solve the following equation for $x$

$$\cos(2x)+\sin(x)=0$$

After simplification i got

$\sin(x)=1$ or $\sin(x)=-\frac{1}{2}$

$\Rightarrow x=90^0$

But don't know how to solve for $x$ for $\sin(x)=-\frac{1}{2}$ ?

Answer 1

If $\space sin(x)=-1/2\space $ then $\space x=\pi n+\frac{\pi}{6}\cdot(-1)^{n+1}$

Edit: $\space n\in Z$

Answer 2

If $\sin x=-1/2$, then $x=210^{\circ}$ or $x=330^{\circ}$ are solutions. In the third and fourth quadrants sine becomes negative. Alternatively you can say $x=-30^{\circ}$ or $x=-150^{\circ}$.

The trick here is to have a mental picture of what the sine function is doing. Also the values are the same when you add or subtract $360^{\circ}$.

The way I was taught to remember which one is positive was remembering a mnemonic for the letters A, S, T and C.

A = All

S = sine

T = tangent

C = cosine

Answer 3

Imagine the unit circle. You should remember from the table of values that $\sin(30^\circ)=\frac12$. By symmetry, $\sin(150^\circ)=\frac12$ too.

Now, the first angle for which $\sin x=-\frac12$ lies in the third quadrant, and by symmetry this angle is $30^\circ+180^\circ=210^\circ$. The second angle lies in the fourth quadrant, and again by symmetry this angle is $150^\circ+180^\circ=330^\circ$.

Answer 4

You can observe that $-\sin x=\cos(\pi/2+x)$, so your equation becomes $$ \cos2x=\cos\left(\frac{\pi}{2}+x\right) $$ which is equivalent to $$ 2x=\frac{\pi}{2}+x+2k\pi \qquad\text{or}\qquad 2x=-\frac{\pi}{2}-x+2k\pi $$ and so, after simplifying, $$ x=\frac{\pi}{2}+2k\pi \qquad\text{or}\qquad x=-\frac{\pi}{6}+2k\frac{\pi}{3} $$ You can now obviate to the “strange” periodicity by dividing the second set of solutions by writing it as $$ x=-\frac{\pi}{6}+2\cdot3k\frac{\pi}{3} \quad\text{or}\quad x=-\frac{\pi}{6}+2\cdot(3k+1)\frac{\pi}{3} \quad\text{or}\quad x=-\frac{\pi}{6}+2\cdot(3k+2)\frac{\pi}{3} $$ (every integer is either a multiple of $3$ or $1$ more than a multiple of $3$ or $2$ more than a multiple of $3$) getting $$ x=-\frac{\pi}{6}+2k\pi \quad\text{or}\quad x=\frac{\pi}{2}+2k\pi \quad\text{or}\quad x=\frac{7\pi}{6}+2k\pi $$ Since these include the other set of solutions, the last sets are the whole sets of solutions.

---

When dealing with $\sin x=-\frac{1}{2}$, you can remember that $\sin(-x)=-\sin x$, so you can write the equation as $$ \sin(-x)=\frac{1}{2} $$ which gives $$ -x=\frac{\pi}{6}+2k\pi \qquad\text{or}\qquad -x=\pi-\frac{\pi}{6}+2k\pi $$ This becomes $$ x=-\frac{\pi}{6}-2k\pi \qquad\text{or}\qquad x=-\frac{5\pi}{6}-2k\pi $$ Since $k$ is supposed to be an arbitrary integer, you can also write the last set of solutions as $$ x=-\frac{\pi}{6}-2(-k-1)\pi \qquad\text{or}\qquad x=-\frac{5\pi}{6}-2(-k-1)\pi $$ so that you get the “principal” solution in the interval $[0,2\pi)$: $$ x=\frac{11\pi}{6}+2k\pi \qquad\text{or}\qquad x=\frac{7\pi}{6}+2k\pi $$

Answer 5

Formally, right from the definition $\sin x:=\frac1{2i}(e^{ix}-e^{-ix})$ we have \begin{align*}\sin x=1&\iff e^{ix}-e^{-ix}=2i\iff e^{2ix}-2ie^{ix}-1=0\iff (e^{ix}-i)^2=0\\ &\iff e^{ix}=i=e^{i\pi/2}\iff\exists k\in\mathbb Z,x=\frac\pi2+2\pi k\end{align*} and similarly \begin{align*}\sin x=-\frac12&\iff e^{ix}+e^{-ix}=-i\iff e^{2ix}-ie^{ix}-1=0\\ &\iff \bigl(e^{ix}-(-i+\sqrt3)/2\bigr)\bigl(e^{ix}-(-i-\sqrt3)/2\bigr)=0\\ &\iff e^{ix}=(-i+\sqrt3)/2=e^{-i\pi/6}\lor e^{ix}=(-i-\sqrt3)/2=e^{-i\,5\pi/6}\\ &\iff\exists k\in\mathbb Z,x=-\frac\pi6+2\pi k\lor x=-\frac{5\pi}6+2\pi k\end{align*}

Note that we've actually solved your original equation in complex numbers, but there were no additional solutions other than the real ones.