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I have always been confused on how to get the other solution when solving trigonometric equations. The original function is $f(x)=8\sin^2(x)-2\sin(x)-4$

$$\sin(x)=\frac{\sqrt{33}\pm1}{8}$$ $$\sin(x)=\frac{\sqrt{33}+1}{8}$$ $$x=\sin^{-1}(\frac{\sqrt{33}+1}{8})$$ $$x=1.00297\\\\\\\\\\\\\\\\\\\\\\\\x=?$$ $$\sin(x)=\frac{\sqrt{33}-1}{8}$$ $$x=\sin^{-1}(\frac{\sqrt{33}-1}{8})$$ $$x=-.634897$$ $$x=?$$ How do I get the other solution pair of solutions? Do I subtract $1.00297$ from $\pi$, because that gets $2.13863$ which I believe is a solution, and if I did the same thing I get $3.77646$ which is a solution too?

3 Answers 3

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Simple. Recall that

$$\sin(x+2\pi k)=\sin(x)\quad k\in\mathbb Z$$

$$\sin(x)=-\sin(-x)$$

And apply this to both of your solutions.

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    $k$ Is any integer. – 2017-01-02
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    I probably am missing the point but isn't $\sin(x+\pi k)=\pm \sin(x)$ depending on the parity of $k$ – 2017-01-02
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    @kingW3 You are right my fine sir. – 2017-01-02
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You can also visualize the solution to equations of the form

$$ \sin\theta=y $$

geometrically with respect to the unit circle. Draw a horizontal line through any value of $y\in[-1,1]$ and note the point(s) of intersection with the unit circle. Any standard angle (vertex the origin, initial side the positive $x$-axis with terminal side containing said point(s) of intersection is a solution of the equation.

For any integer $n$ there will be two corresponding solutions:

  1. $\theta=\phantom{-}\arcsin(y)+2n\pi$
  2. $\theta=-\arcsin(y)+(2n+1)\pi$

Illustration of solutions of sin(x)=y

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Look at this once, enter image description here

I guess you can take it on from here, once by taking positive sign and again by taking negative sign.