If cos⁶x + sin⁶x + k sin²2x = 1, Then Find k
Question:
\[ \cos^6 x+\sin^6 x+k\sin^2 2x=1 \]Find the value of \(k\).
Solution
Use the identity
\[ a^3+b^3=(a+b)^3-3ab(a+b) \]Let
\[ a=\cos^2 x,\qquad b=\sin^2 x \]Then
\[ \cos^6 x+\sin^6 x =(\cos^2 x+\sin^2 x)^3 -3\sin^2 x\cos^2 x(\sin^2 x+\cos^2 x) \] \[ =1-3\sin^2 x\cos^2 x \]Using
\[ \sin 2x=2\sin x\cos x \] \[ \sin^2 x\cos^2 x=\frac{\sin^2 2x}{4} \]Therefore,
\[ \cos^6 x+\sin^6 x = 1-\frac{3}{4}\sin^2 2x \]Substitute into the given equation:
\[ 1-\frac{3}{4}\sin^2 2x +k\sin^2 2x =1 \] \[ \left(k-\frac{3}{4}\right)\sin^2 2x=0 \]Hence,
\[ k=\frac{3}{4} \]