Solve the following trigonometric inequality:

sin2`alpha` sin6`alpha` >=sin3`alpha` sin5`alpha`

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Find all `alpha` such that `sin2alphasin6alpha>=sin3alphasin5alpha` .

(1) This is equivalent to `sin2alphasin6alpha-sin3alphasin5alpha>=0`

(2) Using the identity `sinusinv=1/2[cos(u-v)-cos(u+v)]` we get

`1/2[cos(2alpha-6alpha)-cos(2alpha+6alpha)]-1/2[cos(3alpha-5alpha)-cos(3alpha+5alpha)]>=0`

`1/2[cos(-4alpha)-cos(8alpha)]-1/2[(cos(-2alpha)-cos(8alpha)]>=0`

`1/2cos4alpha-1/2cos8alpha-1/2cos2alpha+1/2cos8alpha>=0` (Using `cos(-alpha)=cosalpha` )

`1/2(cos4alpha-cos2alpha)>=0`

`1/2[cos2(2alpha)-cos2alpha]>=0` Use `cos2u=2cos^2u-1` :

`1/2[2cos^2(2alpha)-cos2alpha-1]>=0`

`1/2[(2cos2alpha+1)(cos2alpha-1)]>=0`

(3) Now `cos2alpha-1<=0` for all `alpha` . So we need `2cos2alpha+1<=0` so that the product is nonnegative.

(4) `2cos2alpha+1<=0`

`cos2alpha<=-1/2` Use `cos2u=2cos^2u-1` :

`2cos^2alpha-1<=-1/2`

`2cos^2alpha<=1/2`

`cos^2alpha<=1/4`

`-1/2<=cosalpha<=1/2`

`cos^(-1)(-1/2)<=alpha<=cos^(-1)(1/2)`

`pi/3<=alpha<=(2pi)/3`

(This is the reference range)

(4)** So the complete solution is all such** `alpha` :

`npi+pi/3<=alpha<=npi+(2pi)/3` for `n in ZZ`

**Sources:**

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