Answer in Trigonometry

We have to use the trigonometric equivalences for the sum of angles and the pitagoric relation to find:

“(sin x+ cos x)^4=Big((sin x+ cos x)^2 Big)^2n\=(sin^2 x+ cos^2 x+2 sin x cos x )^2n \ text{Then we use the relations}n\ sin^2 x+cos^2 x=1n\ text{and } 2 sin x cos x =sin{2x}n \ text{to find}n\ (sin x+ cos x)^4= (1+sin {2x} )^2 n\ (sin x+ cos x)^4= 1+2sin{2x}+sin ^2{2x}”

Then, after we prove that “(sin x+ cos x)^4= 1+2sin{2x}+sin ^2{2x}”, we use one of the relation to find the value of x that safisfies “(sin x+ cos x)^4=0”

“(sin x+ cos x)^4=0n\ (1+sin {2x})^2=0n\ sqrt {(1+2sin {x}cos x)^2}=0n\ 1+2sin {x}cos x=0 n\implies sin {x}cos x=-frac{1}{2}=-frac{1}{sqrt{2}}timesfrac{1}{sqrt{2}}”

Now, since we need a value for x that implies a positive value for the sine and a negative value for the cosine under “90<x<180” and that also has to be equal in magnitude, which implies that x has to be 135° since

“sin{135u00b0}times cos{135u00b0}=cfrac{1}{sqrt{2}}times-cfrac{1}{sqrt{2}}=-cfrac{1}{2}”

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