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Oksanka [162]
3 years ago
7

Help this is for calc

Mathematics
1 answer:
WARRIOR [948]3 years ago
5 0

You can solve for \sin^2\theta, then take square roots:

2\sin^2\theta=1\implies\sin^2\theta=\dfrac12\implies\sin\theta=\pm\dfrac1{\sqrt2}

Then

\theta=\dfrac\pi4+2n\pi,\dfrac{3\pi}4+2n\pi,-\dfrac\pi4+2n\pi,\text{ or }-\dfrac{3\pi}4+2n\pi

where n is any integer; we get the following solutions in the interval 0\le\theta\le2\pi:

\boxed{\theta=\dfrac\pi4,\dfrac{3\pi}4,\dfrac{5\pi}4,\dfrac{7\pi}4}

Or, you can use the double angle identity:

\sin^2\theta=\dfrac{1-\cos(2\theta)}2=\dfrac12\implies1-\cos(2\theta)=1\implies\cos(2\theta)=0

Then

2\theta=\dfrac\pi2+n\pi\implies\theta=\dfrac\pi4+\dfrac{n\pi}2

where n is any integer, and we get the same solutions as above within the prescribed interval.

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