Solve the equation \(\displaystyle{27}{\sin{{\left({x}\right)}}}\cdot{{\cos}^{{2}}{\left({x}\right)}}\cdot{{\tan}^{{3}}{\left({x}\right)}}\cdot{{\cot}^{{4}}{\left({x}\right)}}\cdot{{\sec}^{{5}}{\left({x}\right)}}\cdot{{\csc}^{{6}}{\left({x}\right)}}={256}\)

acidizihvzs

acidizihvzs

Answered question

2022-03-23

Solve the equation
27sin(x)cos2(x)tan3(x)cot4(x)sec5(x)csc6(x)=256

Answer & Explanation

Alannah Farmer

Alannah Farmer

Beginner2022-03-24Added 11 answers

So after reducing the equation the equation will look like that:
27csc6(x)sec2(x)=256
Or:
sin6(x)cos2(x)=27256
Then:
sin6(x)sin8(x)=27256
Or:
sin8(x)sin6(x)+27256=0
I found that sin2(x)34=0 is divisible above equation.I there used the features of high degree polynomials. So the equation will look like in this way:
(sin2(x)34)(sin6(x)+14sin4(x)+34sin2(x)+2764)=0
And:
sin2(x)34=0
The rest is for you.
davane6a7a

davane6a7a

Beginner2022-03-25Added 8 answers

since
27sinxcos2xtan3xcot4xsec5xcsc6x=27sinxcos2xsin3xcos3xcos4xsin4x1cos5x1sin6x
=27cos2xsin6x
so
sin6xcos2x=27256
but on the other hand , we have
t(1t)3=27256
since
256t(1t)327=(4t1)2(16t240t+27)
Because
16t240t+27=(4t5)2+2>0
t=14
then
cosx=12 or 12
where t=sin2x0 then you can do it.

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