Question

cos²x + cos²2x + cos²3x = 1

sin²2x - sin²8x = sin\(\left(\dfrac{17\pi}{2}+10\pi\right)\)

Answer 1

a: \(\Leftrightarrow\dfrac{1+\cos2x}{2}+\dfrac{1+\cos4x}{2}+\dfrac{1+\cos6x}{2}=1\)

\(\Leftrightarrow\cos2x+\cos4x+\cos6x+3=2\)

\(\Leftrightarrow\cos2x+1+\cos4x+\cos6x=0\)

\(\Leftrightarrow2cos^2x-1+1+\cos6x+\cos4x=0\)

\(\Leftrightarrow2cos^2x+cos6x+cos4x=0\)

\(\Leftrightarrow x=\dfrac{\Pi}{2}+k\Pi\)

b: \(\Leftrightarrow\dfrac{1-\cos4x}{2}-\dfrac{1-\cos16x}{2}=1\)

\(\Leftrightarrow\cos16x-\cos4x=2\)

\(\Leftrightarrow x=\pm\dfrac{3}{4}\Pi+k2\Pi\)