202210191050 Solution to 1977-CE-AMATH-II-XX

Prove that

$\displaystyle{\cos^4\theta = \frac{1}{8}(\cos 4\theta + 4\cos 2\theta + 3)}$ .

Hence solve the equation

$\cos 4\theta + 4\cos 2\theta +1=0$

for $0^\circ\leqslant\theta\leqslant 360^\circ$ .

Roughwork.

$\begin{aligned} \textrm{RHS} & = \frac{1}{8}(\cos 4\theta + 4\cos 2\theta + 3) \\ & = \frac{1}{8}\big(\cos 2(2\theta ) + 4\cos 2\theta + 3\big) \\ & = \frac{1}{8}\big( (2\cos^22\theta -1) + 4\cos 2\theta + 3\big) \\ & = \frac{1}{8}(2\cos^22\theta +4\cos 2\theta +2) \\ \dots\,\because\enspace & \cos2\theta = 2\cos^2\theta -1\,\dots \\ & = \frac{1}{8}\big( 2(2\cos^2\theta -1)^2+4(2\cos^2\theta -1)+2\big) \\ & = \dots \\ & = \cos^4\theta \\ & = \textrm{LHS} \\ \end{aligned}$

The missing steps are left the reader.

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