Show that
sin2θsinθ−cos2θcosθ≡secθ \frac{\sin 2\theta}{\sin \theta} - \frac{\cos 2\theta}{\cos \theta} \equiv \sec \theta sinθsin2θ−cosθcos2θ≡secθfor θ≠nπ2\theta \neq \frac{n\pi}{2}θ=2nπ where n∈Zn \in \mathbb{Z}n∈Z.
Solve, for 0∘≤x<45∘0^\circ \le x < 45^\circ0∘≤x<45∘, the equation
7cos2(4x−10∘)=3 7 \cos^2(4x - 10^\circ) = 3 7cos2(4x−10∘)=3giving your answers in degrees to one decimal place. (Solutions based entirely on graphical or numerical methods are not acceptable.)
