# General Solution of Equation cosθ = cosα

## General Solution of Equation cosθ = cosα

General Solution of Equation cosθ = cosα: Given cosθ = cosα

cosθ – cosα = 0

$$\left( \because \cos C-\cos D=2\sin \left( \frac{C+D}{2} \right)\sin \left( \frac{C-D}{2} \right) \right)$$,

cosθ – cosα = $$2\sin \left( \frac{\theta +\alpha }{2} \right)\sin \left( \frac{\theta -\alpha }{2} \right)$$,

$$2\sin \left( \frac{\theta +\alpha }{2} \right)\sin \left( \frac{\theta -\alpha }{2} \right)=0$$,

$$\sin \left( \frac{\theta +\alpha }{2} \right)\sin \left( \frac{\theta -\alpha }{2} \right)=0$$,

$$\sin \left( \frac{\theta +\alpha }{2} \right)=0$$ and

$$\sin \left( \frac{\theta -\alpha }{2} \right)=0$$,

Case(i): $$\sin \left( \frac{\theta +\alpha }{2} \right)=0$$,

$$\left( \frac{\theta +\alpha }{2} \right)=n\pi ,n\in Z$$,

$$\theta +\alpha =2n\pi ,n\in Z$$,

$$\theta =2n\pi -\alpha ,n\in Z$$.

Case(ii): $$\sin \left( \frac{\theta -\alpha }{2} \right)=0$$.

$$\left( \frac{\theta -\alpha }{2} \right)=n\pi ,n\in Z$$,

$$\theta -\alpha =2n\pi ,n\in Z$$,

$$\theta =2n\pi +\alpha ,n\in Z$$,

From case i and ii

$$\theta =2n\pi \pm \alpha ,n\in Z$$,

Note: For general solution of the equation sinθ = k, where -1≤ x ≤ 1. We have cosθ = cos(cos⁻¹k). thus,

θ = 2nπ ± (cos⁻¹k), n ϵ Z.

Example: solve √3 sec2θ = 2

Solution:

√3 sec2θ = 2

sec2θ = 2/√3

2θ = sec⁻¹(2/√3)

2θ = sec⁻¹(sec(π/6))

2θ = π/6

θ = π/12

$$\theta =2n\pi \pm \frac{\pi }{12},n\in Z$$.