- #1

FritoTaco

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## Homework Statement

Find all solutions of the equation in the interval [itex][0, 2\pi][/itex].

[itex]sin6x+sin2x=0[/itex]

## Homework Equations

__Double Angle Formulas__

[itex]sin2x=2sinxcosx[/itex]

[itex]cos2x=cos^{2}x-sin^{2}x [/itex]

[itex]=2cos^{2}x-1 [/itex]

[itex]=1-2sin^{2}x[/itex]

(3 formulas for cos2x)

[itex]=1-2sin^{2}x[/itex]

(3 formulas for cos2x)

[itex]tan2x=\dfrac{2tanx}{1-tan^{2}x}[/itex]

__Sum to Product Formula__

[itex]sinA+sinB=2cos\dfrac{A+B}{2}cos\dfrac{A-B}{2}[/itex]

## The Attempt at a Solution

[itex]sin6x+sin2x=0[/itex]

[itex]2sin\dfrac{6x+2x}{2}cos\dfrac{6x-2x}{2}=0[/itex] (sum to product formula)

[itex]2sin\dfrac{8x}{2}cos\dfrac{4x}{2}[/itex] (simplify)

[itex]2sin4xcos2x=0[/itex]

[itex]2sin2xcos2x[2(cos^{2}x-1)]^{2}=0[/itex] (factor)

It became [itex][2(cos^{2}x-1)]^{2}=0[/itex] because 2sin cancels out and the 4 reduces to 2 which is left with, [itex][2(cos^{2}x-1)]^{2}[/itex] and it's squared because the cos2x is used in the Double Angle Formula. I don't know where to go from here, any suggestions? Thanks.