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I am searching to simplify this expression to find angle values between zero and $2\pi$: $$4\sin^2(x)+\sin(x)-2=0$$

I have been taught to simply the trigonometric expression by factoring or by using identities but the way to proceed on this one escapes me. Otherwise I could find the answer to an equation such as the following by factoring in the following way:

$$ \begin{align*} 0 &= 2\cos^2(x)+\cos(x)-1 \\ &= (2\cos(x)-1)(\cos(x)+1) \\ \end{align*}\\ \implies \cos(x)=\frac{1}{2} \quad\text{or}\quad \cos(x)=-1 $$ So the angle $x$ would be $\pi/3$, $5\pi/3$, or $\pi$ ... in the range of $0

I cannot factor the expression I have provided in bold by doing this process. Any recommendations?

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    If you make the substitution $y=sin(x)$, then it becomes $ 4y^2+y-2=0$. It's not gonna be very clean – 2017-02-08
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    How did you get from sines in the original equation to cosines in the next line, Bernardo? – 2017-02-08
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    Even though the equation does not factor nicely, you still obtain two real solutions for $\sin(x)$ in the range $[-1,1]$ so there will be four solutions for $x$ in $[0,2\pi]$. It's just that you will have to use the arcsine function to find them. Now if that had been $4\sin^2(x)+\sin(x)-3=0\cdots$ – 2017-02-08

2 Answers 2

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You should introduce the substitution: $\sin{x}=t$.

Now your equation looks like this: $4t^2+t-2=0$.

You can solve it as a regular quadratic equation.

$$t_{1,2}=\frac{-1\pm\sqrt{33}}{8}$$

Then you ought to revert the substitution.

$$t_1=\sin{x_1}=\frac{-1-\sqrt{33}}{8}\Rightarrow x_1=\arcsin{\frac{-1-\sqrt{33}}{8}}$$ $$t_2=\sin{x_2}=\frac{-1+\sqrt{33}}{8}\Rightarrow x_2=\arcsin{\frac{-1+\sqrt{33}}{8}}$$

Since you require $x\in(0, 2\pi)$ and $x_1$ is negative, you can add a period to the solution and get:

$$x_1=2\pi + \arcsin{\frac{-1-\sqrt{33}}{8}}$$ $$x_2=\arcsin{\frac{-1+\sqrt{33}}{8}}$$

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It's just $$4\sin^2x+\sin x-2=4\left(\sin x-\frac{-1-\sqrt{33}}{8}\right)\left(\sin x-\frac{-1+\sqrt{33}}{8}\right)$$