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HELP PLS ASAP

1/2 sin x sin (2x) + cos^3 x

User Jaanu
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1 Answer

1 vote

Answer:

x=π/8,5π/8,2π/3,9π/8,4π/3,13π/8

Explanation:

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

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.

User Redsquare
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6.9k points