# General Solution of Equation cos²θ = cos²α

## General Solution of Equation cos²θ = cos²α

General Solution of Equation cos²θ = cos²α (or) sin²θ = sin²α:

cos²θ = cos²α

(since cos²θ + sin²θ = 1 ⇒ cos²θ = 1 – sin²θ)

(1 – sin²θ) = (1 – sin²α)

(1 – sin²θ) – (1 – sin²α) = 0

sin²θ = sin²α

sin²θ – sin²α = 0

sin (θ + α) sin (θ – α) = 0

Case(i):

sin (θ + α) = 0

θ + α = sin⁻¹ (0)

θ + α = nπ, n ϵ Z

θ = nπ – α, n ϵ Z

Case(ii):

sin (θ – α) = 0

θ – α = sin⁻¹(0)

θ – α = nπ, n ϵ Z

θ = nπ + α, n ϵ Z

From case(i) and (ii)

θ = nπ ± α, n ϵ Z

Example: Solve $${{\log }_{\tan x}}\left( 2+4{{\cos }^{2}}x\right)=2$$.

Solution:

$${{\log }_{\tan x}}\left( 2+4{{\cos }^{2}}x \right)=2$$,

$$\left( 2+4{{\cos }^{2}}x \right)={{\tan }^{2}}x$$,

$$2+4{{\cos }^{2}}x=-1+{{\sec }^{2}}x$$,

$$3+4{{\cos }^{2}}x={{\sec }^{2}}x$$,

$$3+4{{\cos }^{2}}x=\frac{1}{{{\cos }^{2}}x}$$,

$$3{{\cos }^{2}}x+4{{\cos }^{4}}x=1$$,

$$3{{\cos }^{2}}x+4{{\cos }^{4}}x-1=0$$,

$$4{{\cos }^{4}}x+3{{\cos }^{2}}x-1=0$$,

$$\left( 4{{\cos }^{2}}x-1 \right)\left( {{\cos }^{2}}x+1 \right)=0$$,

$$\left( {{\cos }^{2}}x+1 \right)=0$$,

$${{\cos }^{2}}x=-1$$,

$$\left( 4{{\cos }^{2}}x-1 \right)=0$$,

$$4{{\cos }^{2}}x=1\Rightarrow {{\cos }^{2}}x=\frac{1}{4}$$,

$${{\cos }^{2}}x={{\cos }^{2}}\frac{\pi }{3}$$,

The general solution is x = nπ ± π/3, n ϵ I.