Question 138824

Let *[Tex \LARGE u=\frac{\theta}{2}]. So this means that *[Tex \LARGE 2u=2*\frac{\theta}{2}=\theta]

*[Tex \LARGE \sin^2\left(u\right)-\sin^2\left(2u\right)=0  ] Replace *[Tex \LARGE \frac{\theta}{2}] with u and *[Tex \LARGE \theta] with 2u

*[Tex \LARGE \sin^2\left(u\right)-\left(\sin\left(2u\right)\right)^2=0  ] Rewrite *[Tex \LARGE \sin^2\left(2u\right)] as *[Tex \LARGE \left(\sin\left(2u\right)\right)^2]

*[Tex \LARGE \sin^2\left(u\right)-\left(2\sin\left(u\right)\cos\left(u\right)\right)^2=0 ] Replace *[Tex \LARGE \sin\left(2u\right)] with *[Tex \LARGE 2\sin\left(u\right)\cos\left(u\right)]

*[Tex \LARGE \sin^2\left(u\right)-4\sin^2\left(u\right)\cos^2\left(u\right)=0 ] Square *[Tex \LARGE 2\sin\left(u\right)\cos\left(u\right)]

*[Tex \LARGE \sin^2\left(u\right)\left(1-4\cos^2\left(u\right)\right)=0 ] Factor out *[Tex \LARGE \sin^2\left(u\right)]

Now use the zero product property:

*[Tex \LARGE \sin^2\left(u\right)=0]  ...or...  *[Tex \LARGE 1-4\cos^2\left(u\right)=0 ]

Now let's solve *[Tex \LARGE \sin^2\left(u\right)=0]:

*[Tex \LARGE \sin^2\left(u\right)=0 ]

*[Tex \LARGE \sin\left(u\right)=0 ] Take the square root of both sides

*[Tex \LARGE u=0]  ...or...  *[Tex \LARGE u=\pi ] Take the arcsine of both sides

Since *[Tex \LARGE u=\frac{\theta}{2}], this means

*[Tex \LARGE \theta=0]  ...or...  *[Tex \LARGE \theta=2\pi ]

However, since  *[Tex \LARGE 2\pi ] is not in the interval [0,2*[Tex \large \pi ]), it is not a solution.

So the first solution is *[Tex \LARGE \theta=0]

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Now let's solve *[Tex \LARGE  1-4\cos^2\left(u\right)=0 ]:

*[Tex \LARGE  1-4\cos^2\left(u\right)=0 ]

*[Tex \LARGE  -4\cos^2\left(u\right)=-1 ] Subtract 1 from both sides

*[Tex \LARGE  \cos^2\left(u\right)=\frac{1}{4} ] Divide both sides by -4

Take the square root of both sides:

*[Tex \LARGE  \cos\left(u\right)=\frac{1}{2}]  ...or...  *[Tex \LARGE \cos\left(u\right)=-\frac{1}{2} ]

So let's solve the first part *[Tex \LARGE \cos\left(u\right)=\frac{1}{2} ]:

*[Tex \LARGE \cos\left(u\right)=\frac{1}{2} ]

*[Tex \LARGE u=\frac{\pi}{3}]  ...or...  *[Tex \LARGE u=-\frac{\pi}{3} ]

*[Tex \LARGE \theta=\frac{2\pi}{3}]  ...or...  *[Tex \LARGE \theta=-\frac{2\pi}{3} ]

However since our interval is positive, the negative answer is not in the interval.

So another part of the solution is *[Tex \LARGE \theta=\frac{2\pi}{3}]

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Now let's solve the second part *[Tex \LARGE \cos\left(u\right)=-\frac{1}{2} ]:

*[Tex \LARGE \cos\left(u\right)=-\frac{1}{2} ]

*[Tex \LARGE u=\frac{2\pi}{3}]  ...or...  *[Tex \LARGE u=-\frac{2\pi}{3} ]

*[Tex \LARGE \theta=\frac{4\pi}{3}]  ...or...  *[Tex \LARGE \theta=-\frac{4\pi}{3} ]

So another part of the solution is *[Tex \LARGE \theta=\frac{4\pi}{3}]

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*[Tex \LARGE \theta=0,\theta=\frac{2\pi}{3}, \hspace {12} \textrm{or} \hspace {12} \theta=\frac{4\pi}{3}]