Question

a1" title="(cos^{2} x-sin^{2}x)^{2} -sin4x+sin^{2}2x=0" alt="(cos^{2} x-sin^{2}x)^{2} -sin4x+sin^{2}2x=0" align="absmiddle" class="latex-formula">

Answer 1

Answer:

x=22.5°

Step-by-step explanation:

(cos²x-sin²x)-sin4x+sin²2x=0

or, (cos2x)²-sin4x+sin²2x=0

or, cos²2x+sin²2x-sin4x=0

or, 1-sin4x=0

or, sin4x=1

or, 4x=90°

or, x=90°/4

or, x=22.5°