θ∈(−2π,2π)
θ=5nπ,n=0,±1,±2
tanθ=cot5θ
sin2θ=cos4θ
To find: the number of such θ
1. Solve tanθ=cot5θ
2. Solve sin2θ=cos4θ
3. Take common values, drop θ=5nπ
cotA=tan(2π−A)
cot5θ=tan(2π−5θ)
tanθ=tan(2π−5θ)
θ=2π−5θ+kπ
6θ=2π+kπ
θ=12π+6kπ
k=0: 12π;k=1: 4π;k=2: 125π
k=−1: −12π;k=−2: −4π;k=−3: −125π
θ∈{±12π, ±4π, ±125π}
⇒6 values from the first equation\cos 2x = 1 - 2\sin^2 x\cos 4\theta = 1 - 2\sin^2 2\theta\sin 2\theta = 1 - 2\sin^2 2\theta2\sin^2 2\theta + \sin 2\theta - 1 = 02s^2 + s - 1 = 0,\quad s = \sin 2\theta2s^2 + s - 1 = (2s - 1)(s + 1) = 0\sin 2\theta = \dfrac{1}{2} \quad \text{or} \quad \sin 2\theta = -12\theta = \dfrac{\pi}{6} + 2k\pi \ \text{ or } \ \dfrac{5\pi}{6} + 2k\pi\theta = \dfrac{\pi}{12} + k\pi \ \text{ or } \ \dfrac{5\pi}{12} + k\pi\theta = \dfrac{\pi}{12},\ \dfrac{5\pi}{12}2\theta = -\dfrac{\pi}{2} + 2k\pi\theta = -\dfrac{\pi}{4} + k\pi \ \Rightarrow\ \theta = -\dfrac{\pi}{4}\sin 2\theta = \cos 4\theta \Rightarrow \theta = -\dfrac{\pi}{4},\ \dfrac{\pi}{12},\ \dfrac{5\pi}{12}\text{Common: } \theta = -\dfrac{\pi}{4},\ \dfrac{\pi}{12},\ \dfrac{5\pi}{12}\text{Excluded: } \theta = 0,\ \pm\dfrac{\pi}{5},\ \pm\dfrac{2\pi}{5}-\dfrac{\pi}{4},\ \dfrac{\pi}{12},\ \dfrac{5\pi}{12} \ \notin\ \left{0,\ \pm\dfrac{\pi}{5},\ \pm\dfrac{2\pi}{5}\right}\text{Number of values of } \theta = 3$