Cos 4θ − cos 2θ = sin θ

1 answer:

7 0

$cos(4\theta) - cos(2\theta) = sin(\theta)$
$-2sin(\frac{4\theta + 2\theta}{2})sin(\frac{4\theta - 2\theta}{2}) = sin(\theta)$
$-2sin(\frac{6\theta}{2}sin\frac{2\theta}{2}) = sin(\theta)$
$-2sin(3\theta)sin(\theta) = sin(\theta)$
$\frac{-2sin(3\theta)sin(\theta)}{-2sin(\theta)} = \frac{sin(\theta)}{-2sin(\theta)}$
$sin(3\theta) = -\frac{1}{2}$
$sin^{-1}[sin(3\theta)] = sin^{-1}(-\frac{1}{2})$
$3\theta = -30$
$\frac{3\theta}{3} = \frac{-30}{3}$
$\theta = -10$

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