Question

a. Sin mũ 4 x - cos mũ 4 x =1

b. Sin mũ 4 x + cos mũ 4 x = 1

Answer 1

a/ \(sin^4x-cos^4x=1\)

\(\Leftrightarrow\left(sin^2x+cos^2x\right)\left(sin^2x-cos^2x\right)=1\)

\(\Leftrightarrow-cos2x=1\)

\(\Rightarrow cos2x=-1\)

\(\Rightarrow2x=\pi+k2\pi\)

\(\Rightarrow x=\frac{\pi}{2}+k\pi\)

b/ \(sin^4x+cos^4x=1\)

\(\Leftrightarrow\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x=1\)

\(\Leftrightarrow sin^2x.cos^2x=0\)

\(\Leftrightarrow sin2x=0\)

\(\Rightarrow2x=k\pi\Rightarrow x=\frac{k\pi}{2}\)