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MarkFL

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We are given to solve:

$$\displaystyle 3\sin^2(x)-2\sin(x)-3=0$$ where $$\displaystyle 0<x<2\pi$$

Recognizing that we have a quadratic in $\sin(x)$, we may use the quadratic formula to state:

$$\displaystyle \sin(x)=\frac{-(-2)\pm\sqrt{(-2)^2-4(3)(-3)}}{2(3)}=\frac{2\pm\sqrt{40}}{6}=\frac{1\pm\sqrt{10}}{3}$$

Since we require $$\displaystyle -1\le\sin(x)\le1$$ we discard the positive root, and we are left with:

$$\displaystyle \sin(x)=\frac{1-\sqrt{10}}{3}$$

Hence:

$$\displaystyle x=\sin^{-1}\left(\frac{1-\sqrt{10}}{3} \right)$$

Since this is less than zero, we need to add $$\displaystyle 2\pi$$ to get the equivalent angle in the required interval:

$$\displaystyle x=2\pi+\sin^{-1}\left(\frac{1-\sqrt{10}}{3} \right)$$

Now, this is the 4th quadrant solution, but we should observe there is also a 3rd quadrant solution, given by:

$$\displaystyle x=\pi-\sin^{-1}\left(\frac{1-\sqrt{10}}{3} \right)$$

Note: this comes from the identity $$\displaystyle \sin(\pi-\theta)=\sin(\theta)$$.

If we are to use a calculator to obtain decimal approximations, then:

$$\displaystyle x=2\pi+\sin^{-1}\left(\frac{1-\sqrt{10}}{3} \right)\approx5.47828834852818$$

$$\displaystyle x=\pi-\sin^{-1}\left(\frac{1-\sqrt{10}}{3} \right)\approx3.9464896122412$$

To answerer and any other guests viewing this topic, I invite and encourage you to post other trigonometry questions here in our Trigonometry forum.

Best Regards,

Mark.