Hi please help me thanksss
pls provide workings too thanks :)

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Class 11

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>>Solve for x: sin x + sin 2x...

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Solve for x:sinx+sin2x+sin3x=cosx+cos2x+cos3x in the interval 0≤x≤2π.

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Solution

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We write the given equation as

(sinx+sin3x)+sin2x=(cosx+cos3x)+cos2x

or 2sin2xcosx+sin2x=2cos2xcosx+cos2x

or sin2x(2cosx+1)=cos2x(2cosx+1)

or (sin2x−cos2x)(2cosx+1)=0

∴sin2x−cos2x=0 or 2cosx+1=0

If sin2x−cos2x=0, then tan2x=1,

Hence 2x=nπ+π/4

or x=(4n+1)

8

π

.(1)

If 2cosx+1=0, then cosx=−1/2 (2)

∴x=2nπ±

3

2π

or x=

3

6n±2

π

We seek values of x in the interval 0≤x≤2π

In this interval (1) gives

x=π/8,5π/8,9π/8,13π/8. (n=0,1,2,3)

and (2) gives x=2π/3,4π/3. (for n=0,1)

Thus we get the answer

x=π/8,5π/8,2π/3,9π/8,4π/3,13π/8.

215.92 dollars is the answer