Question

Find all solutions in the interval [0, 2π). 2 sin2x = sin x

Answer 1

Answer:

The solutions on the given interval are :

$0$

$\pi$

$\cos^{-1}(\frac{1}{4})$

$-\cos^{-1}(\frac{1}{4})+2\pi$

Step-by-step explanation:

We will need the double angle identity $\sin(2x)=2\sin(x)\cos(x)$.

Let's begin:

$2\sin(2x)=\sin(x)$

Use double angle identity mentioned on left hand side:

$2\cdot 2\sin(x)\cos(x)=\sin(x)$

Simplify a little bit on left side:

$4\sin(x)\cos(x)=\sin(x)$

Subtract $\sin(x)$ on both sides:

$4\sin(x)\cos(x)-\sin(x)=0$

Factor left hand side:

$\sin(x)[4\cos(x)-1]=0$

Set both factors equal to 0 because at least of them has to be 0 in order for the equation to be true:

$\sin(x)=0 \text{ or } 4\cos(x)-1=0$

The first is easy what angles $\theta$ are $y$-coordinates on the unit circle 0. That happens at $0$ and $\pi$ on the given range of $x$ (this $x$ is not be confused with the $x$-coordinate).

Now let's look at the second equation:

$4\cos(x)-1=0$

Isolate $\cos(x)$.

Add 1 on both sides:

$4 \cos(x)=1$

Divide both sides by 4:

$\cos(x)=\frac{1}{4}$

This is not as easy as finding on the unit circle.

We know $\arccos( )$ will render us a value between $0$ and $2\pi$.

So one solution on the given interval for x is $x=\cos^{-1}(\frac{1}{4})$.

We know cosine function is even.

So an equivalent equation is:

$\cos(-x)=\frac{1}{4}$

Apply $\cos^{-1}$ to both sides:

$-x=\cos^{-1}(\frac{1}{4})$

Multiply both sides by -1:

$x=-\cos^{-1}(\frac{1}{4})$

This going to be negative in the 4th quadrant but if we wrap around the unit circle, $2\pi$ , we will get an answer between $0$ and $2\pi$.

So the solutions on the given interval are :

$0$

$\pi$

$\cos^{-1}(\frac{1}{4})$

$-\cos^{-1}(\frac{1}{4})+2\pi$