Answer
The solution set is
$$\{0.8751+2\pi n, 2.2665+2\pi n,3.5908+2\pi n, 5.834+2\pi n, n\in Z\}$$
Work Step by Step
$$\sin x(3\sin x-1)=1$$
1) Solve the equation over the interval $[0,2\pi)$
$$\sin x(3\sin x-1)=1$$
$$3\sin^2x-\sin x-1=0$$
Consider the equation as a quadratic formula, with $a=3, b=-1, c=-1$
- Calculate $\Delta$: $\Delta=b^2-4ac=(-1)^2-4\times3\times(-1)=1+12=13$
- Find out $\sin x$: $$\sin x=\frac{-b\pm\sqrt\Delta}{2a}=\frac{1\pm\sqrt{13}}{6}$$
- For $\sin x=\frac{1+\sqrt{13}}{6}$. $x$ would be
$$x=\sin^{-1}\frac{1+\sqrt{13}}{6}$$
$$x\approx0.8751$$ (Be careful with degrees and radians. Here we need to use radians.)
However, over the interval $[0,2\pi)$, there is one more value of $x$ where $\sin x=\frac{1+\sqrt{13}}{6}$, which is $x=\pi-0.8751\approx2.2665$
In total, $x\in\{0.8751, 2.2665\}$
- For $\sin x=\frac{1-\sqrt{13}}{6}$. $x$ would be
$$x=\sin^{-1}\frac{1-\sqrt{13}}{6}$$
$$x\approx-0.4492$$
In the interval $[0,2\pi)$, that equals to $x\approx-0.4492+2\pi\approx5.834$
However, over the interval $[0,2\pi)$, there is one more value of $x$ where $\sin x=\frac{1-\sqrt{13}}{6}$, which is $x=\pi-(-0.4492)\approx3.5908$
In total, $x\in\{3.5908, 5.834\}$
Therefore, overall, $$x=\{0.8751, 2.2665,3.5908, 5.834\}$$
2) Solve the equation for all solutions
- The integer multiples of the period of the sine function is $2\pi$.
- We apply it to each solution found in step 1. The results are the solution set.
So the solution set is
$$\{0.8751+2\pi n, 2.2665+2\pi n,3.5908+2\pi n, 5.834+2\pi n, n\in Z\}$$
