# find what values of x accomplishes:sin (pi cos^2 x) + sin (`pi` sin^2 x) = 1

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Would it go better:

sin(`pi` cos^2 x) + sin(`pi` sin^2 x)=

=2 sin (`pi` /2) cos(`pi` cos2x) (formula Prosthaphaeresis)

so: 2 cos(`pi` cos2x)=1 cos( `pi` cos2x)= 1/2

`pi` cos 2x =`pi` /3 `pi` cos 2x= 5`pi` /6

cos2x = 1/3 cos2x = 5/6

x = 40° 12' 10'' x = 33°33'26'' ?

Let `cos^2x=y` , so `sin^2x=1-y`

thus question

`sin(pi.cos^2(x))+sin(pi.sin^2(x))=1` ,reduces to

`sin(pi.y)+sin(pi.(1-y))=1`

`sin(pi.y)+sin(pi-pi.y)=1`

`sin(pi.y)+sin(pi.y)=1`

`2sin(pi.y)=1`

`sin(pi.y)=1/2`

`sin(pi.y)=sin(pi/6)`

`pi.y=2npi+(-1)^n(pi/6)`

`y=2n+(-1)^n(1/6)`

`` where n is an integer ,let n=o

y=`1/6`

thus

`cos^2(x)=1/6`

`cos(x)=+-sqrt(1/6)`

x=cos^(-1)(sqrt(1/6))

`x=cos^(-1)(sqrt(1/6))`

`x=69.91 `

We have taken positive only due to inverse restriction.