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marysya [2.9K]
3 years ago
14

Save me the headache

Mathematics
2 answers:
vichka [17]3 years ago
8 0

Answer:

Simple tell everyone around you to be quiet and to keep quiet

if u have a headache

Step-by-step explanation:

maxonik [38]3 years ago
5 0

(9\sin2x+9\cos2x)^2=81

Taking the square root of both sides gives two possible cases,

9\sin2x+9\cos2x=9\implies\sin2x+\cos2x=1

or

9\sin2x+9\cos2x=-9\implies\sin2x+\cos2x=-1

Recall that

\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta

If \alpha=2x and \beta=\dfrac\pi4, we have

\sin\left(2x+\dfrac\pi4\right)=\dfrac{\sin2x+\cos2x}{\sqrt2}

so in the equations above, we can write

\sin2x+\cos2x=\sqrt2\sin\left(2x+\dfrac\pi4\right)=\pm1

Then in the first case,

\sqrt2\sin\left(2x+\dfrac\pi4\right)=1\implies\sin\left(2x+\dfrac\pi4\right)=\dfrac1{\sqrt2}

\implies2x+\dfrac\pi4=\dfrac\pi4+2n\pi\text{ or }\dfrac{3\pi}4+2n\pi

(where n is any integer)

\implies2x=2n\pi\text{ or }\dfrac\pi2+2n\pi

\implies x=n\pi\text{ or }\dfrac\pi4+n\pi

and in the second,

\sqrt2\sin\left(2x+\dfrac\pi4\right)=-1\implies\sin\left(2x+\dfrac\pi4\right)=-\dfrac1{\sqrt2}

\implies2x+\dfrac\pi4=-\dfrac\pi4+2n\pi\text{ or }-\dfrac{3\pi}4+2n\pi

\implies2x=-\dfrac\pi2+2n\pi\text{ or }-\pi+2n\pi

\implies x=-\dfrac\pi4+n\pi\text{ or }-\dfrac\pi2+n\pi

Then the solutions that fall in the interval [0,2\pi) are

x=0,\dfrac\pi4,\dfrac\pi2,\dfrac{3\pi}4,\pi,\dfrac{5\pi}4,\dfrac{3\pi}2,\dfrac{7\pi}4

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