Solve trig-equation

Question

Can someone solve the equation $3\sin(x) + 2\sin^2(x) = 1$? I know that one answer is $x = 1/2\pi + k\cdot2\pi$ with $k \in \mathbb{Z}$ but can't seem to find the other one.

Answer 1

$$2 \sin^2 x+3 \sin x-1=0$$

For a while, let $\sin x$ is $y$,

$$2y^2+3y-1=0$$ Now, $$y=\frac{-3\pm\sqrt{9+8}}{4}$$ $$y=\frac{-3}{4}\pm\frac{\sqrt{17}}{4}$$

Now, ignore the $-$ part because it goes out of $\sin$ range which is $\in[-1,1]$

So, you must consider $y=\frac{-3}{4}+\frac{\sqrt{17}}{4}$.

Which happens to be unsolvable in normal conventions (Not easily reducable in conventions you have written)

Answer 2

Never mind the old question. It was wrong; the plus sign had to be a minus sign, hence the right equation is: $$3\sin(x)-2\sin^2(x) = 1$$

I know from Wolfram Alpha that the answers in domain $[0, 2\pi]$ are $x = \pi/2 \;\vee x= \pi/6 \; \vee x = 5\pi/6$, but how can I algebraically show this?

Answer 3

To answer the new question posted in the comments: $$3\sin{x}-2\sin^{2}{x}=1$$ Let: $$\sin{(x)}=\beta$$ Thus: $$-2\beta^{2}+3\beta=1$$ Dividing everything by -2 $$\beta^{2}-\frac{3}{2}\beta=-\frac{1}{2}$$ Completing the square: $$(\beta-\frac{3}{4})^{2}=\frac{1}{16}$$ Taking the square root: $$\beta-\frac{3}{4}=\pm\frac{1}{4}$$ $$\beta=\pm\frac{1}{4}+\frac{3}{4}$$ $$\beta=1$$ $$\beta=\frac{1}{2}$$ Solving for sin(x): $$1=\sin{x}$$ $$\frac{1}{2}=\sin{x}$$ Recognizably the sin(90)=1, therefore, x=90. Recognizing sin(30)=.5, thus, x=30. And the reference angle of 150 is 30, and as such$$\sin{150}=\sin{30}$$ So your solutions are:
x=90
x=30
x=Any number with a reference angle of 30, or 90, which includes 150.
In radians: $$\mathit{x}=90=\frac{\pi}{2}$$ $$\mathit{x}=30=\frac{\pi}{6}$$