Find all solutions exactly for $2\sin^2 x - \sin x = 0$

Question

How can I solve this Question ?

• Find all solutions exactly for $2\sin^2 x - \sin x = 0$

my answer was $( 2k\pi,\pi+2k\pi, 11\pi/6, 7\pi/6)$ but my teacher answer was $( 2k\pi,\pi+2k\pi, \pi/6 +2k\pi, 5\pi/6 +2k\pi )$

Answer 1

HINT: solve the equation $$\sin(x)(2\sin(x)-1)=0$$

Answer 2

Hint: $$2\sin^2x-\sin x = 0 \implies \color{red}{2}\sin^2x + \color{blue}{(-1)}\sin x + \color{green}{0} = 0,$$ so we have $$\sin x = \frac{-\color{blue}{(-1)}\pm \sqrt{\color{blue}{(-1)}^2 - 4\cdot \color{red}{2}\cdot\color{green}{0}}}{2\cdot \color{red}{2}} = \frac{1 \pm 1}{4} = \begin{cases}\frac{1}{2} \\ 0\end{cases}.$$

See when $\sin x = 1/2$ or $0$ and consider all possible values.

Answer 3

You have $2 \sin^2 x - \sin x =0$ or $\sin x ( 2 \sin x -1) =0$, so $\sin x = 0$ or $2 \sin x -1 = 0$.

$\sin x =0$ implies $x = \pi k$ for integer $k$.

$2 \sin x - 1 =0 $ means $\sin x = \frac{1}{2}$ which has solutions between $0, 2pi$ as $\pi/6$ and $5 \pi/6$. Since $\sin$ is $2 \pi$ periodic, you get $\pi/6 + 2 \pi k$ and $5 \pi/6 + 2 \pi k$ for integer $k$.

Answer 4

We have solutions to $2y^2-y=0$ as $$y = \frac{1 \pm \sqrt{1-0}}{4} \implies y = 0,\frac{1}{2}$$ With each pass around the unit circle we know $y = \sin(x)$ takes the value $0$ twice, and the value $\frac{1}{2}$ twice. Can you see why your teacher is correct now?

Answer 5

We have $$2\sin^2 x-\sin x=0 $$ $$\implies \sin x(2\sin x-1)=0$$ We have $$\sin x=0 \implies \color{blue}{x=n\pi}$$ Where, $n$ is any integer & $$2\sin x-1=0 $$$$\implies \sin x=\frac{1}{2}=\sin\frac{\pi}{6} \implies \color{blue}{x=2n\pi+\frac{\pi}{6}}\quad \text{&} \quad \color{blue}{x=2n\pi+\frac{5\pi}{6}}$$ Now, writing the complete solution for $x$, we have $$\color{blue}{x\in \{n\pi\}\cup \{2n\pi+\frac{\pi}{6} \}\cup\{2n\pi+\frac{5\pi}{6}\}}$$ Where, $n$ is any integer

Answer 6

Check your solution:

$$\sin\left(\frac{11\pi}6\right)=\sin\left(\frac{7\pi}6\right)=-\frac12\to2\frac14+\frac12\color{red}{=0}.$$

And the Teacher's solution:

$$\sin\left(\frac{\pi}6\right)=\sin\left(\frac{5\pi}6\right)=\frac12\to2\frac14-\frac12\color{green}{=0}.$$