You may use the identity $cos⁡3θ≡4cos⁡3θ−3cos⁡θ\cos 3\theta \equiv 4\cos^3\theta - 3\cos\theta$ .

Show that the equation $1 - \cos 3x = \sin^2 x$ can be written as

$cos⁡x(4cos⁡x+3)(cos⁡x−1)=0\cos x\left(4\cos x + 3\right)\left(\cos x - 1\right) = 0$

Hence solve, for $−π≤x≤π-\pi \leq x \leq \pi$ , the equation

$1 - \cos 3x = \sin^2 x$

Give your answers to three significant figures where they are not exact.

You may use the identity $cos⁡3θ≡4cos⁡3θ−3cos⁡θ\cos 3\theta \equiv 4\cos^3\theta - 3\cos\theta$ .

Show that the equation $1 - \cos 3x = \sin^2 x$ can be written as

$cos⁡x(4cos⁡x+3)(cos⁡x−1)=0\cos x\left(4\cos x + 3\right)\left(\cos x - 1\right) = 0$

Hence solve, for $−π≤x≤π-\pi \leq x \leq \pi$ , the equation

$1 - \cos 3x = \sin^2 x$

Give your answers to three significant figures where they are not exact.

1 Pure Mathematics

1 Pure Mathematics Questions