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.