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9sin(2x)=9cos(x) find x.

Sagot :

Answer:

[tex]x=\frac{\pi}{2}+\pi n, \frac{\pi}{6}+2\pi n, \frac{5\pi}{6}+2\pi n[/tex]

Step-by-step explanation:

[tex]9sin(2x)=9cos(x)[/tex] <-- Starting Equation

[tex]9sin(2x)-9cos(x)=0[/tex] <-- Move all terms to the left and set equal to 0

[tex]9(sin(2x)-cos(x))=0[/tex] <--Simplify

[tex]9(2sin(x)cos(x)-cos(x))=0[/tex] <-- Trig Identity (sin2x=2sinxcosx)

[tex]9(cos(x)(2sin(x)-1)=0[/tex] <-- Simplify

[tex]9cos(x)(2sin(x)-1)=0[/tex] <-- Simplify

Use Zero Product Property:

[tex]9cos(x)=0[/tex] and [tex]2sin(x)-1=0[/tex]

Use the unit circle to find your solutions:

[tex]x=\frac{\pi}{2}+\pi n, \frac{\pi}{6}+2\pi n, \frac{5\pi}{6}+2\pi n[/tex]

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