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iragen [17]
3 years ago
5

Find all solutions to the equation in the interval [0, 2TT). Cos x= sin 2x (1 point)

Mathematics
1 answer:
igomit [66]3 years ago
6 0

You can use the formula for the double angle of the sine to write

\sin(2x)=2\sin(x)\cos(x)

To turn the equation into

2\sin(x)\cos(x)=\cos(x)

Move everything to the left hand side to have

2\sin(x)\cos(x)-\cos(x)=0

Factor \cos(x) to have

\cos(x)(2\sin(x)-1)=0

A factor equals zero if and only if one of its factor equals zero. So, either

\cos(x)=0

or

2\sin(x)-1=0 \iff 2\sin(x)=1\iff \sin(x)=\dfrac{1}{2}

I assume that you have a table to lookup for these known values. If you have troubles solving for x, hit me up in the comments

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Given data:

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<em>If a tangent and a secant intersect at the exterior of a circle then the measure of angle formed is one-half the positive difference of the measures of the intercepted arcs.</em>

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\qquad\qquad\huge\underline{{\sf Answer}}

\textbf{Since it's a right angled Triangle we can }  \\ \textbf{ use Pythagoras theorem }

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