General Solution of Some Standard Equations
General solution of Equation sinθ = sinα:
sinθ = sinα
sinθ – sinα = 0,
\(\left( \because \ \sin C-\sin D=2\cos \left( \frac{C+D}{2} \right)\sin \left( \frac{C-D}{2} \right) \right)\),
\(2\cos \left( \frac{\theta +\alpha }{2} \right)\sin \left( \frac{\theta -\alpha }{2} \right)=0\),
\(\cos \left( \frac{\theta +\alpha }{2} \right)=0\),
\(\left( \frac{\theta +\alpha }{2} \right)={{\cos }^{-1}}\left( 0 \right)\),
\(\left( \frac{\theta +\alpha }{2} \right)=\left( 2m+1 \right)\frac{\pi }{2}\),
θ + α = (2m + 1) π
θ = (2m + 1) π – α, m ϵ Z
\(\theta \text{ }=\text{ }\left( 2m\text{ }+\text{ }1 \right)\text{ }\pi +{{\left( -1 \right)}^{2m+1}}\text{ }\alpha ,\text{ }m\text{ }\epsilon \text{ }Z\)….(1)
\(\sin \left( \frac{\theta -\alpha }{2} \right)=0\)(θ – α)/2 = sin⁻¹(0)
(θ – α)/2 = mπ
(θ – α) = 2mπ
θ = 2mπ + α m ϵ Z
\(\theta ~=\text{ }2m\pi \text{ }+\text{ }{{\left( -1 \right)}^{2m}}\alpha ,\ \ ~m\text{ }\epsilon \text{ }Z\)…(2)
From equation (1) and (2) combining
\(\theta =n\pi +{{\left( -1 \right)}^{n}}\alpha ,\ n\in Z\),
Note: For general solution of the equation sinθ = k, where -1 ≤ k ≤ 1.we have \(sin\theta \text{ }=\text{ }sin(si{{n}^{-1}}k)\) then \(\theta \text{ }=n\pi +{{\left( -1 \right)}^{n}}(si{{n}^{-1}}k),\ n\in Z\).
Example: Solve sin³θ cosθ – cos³θsinθ = ¼
solution:
sin³θ cosθ – cos³θsinθ = ¼
4sinθ cosθ(sin²θ – cos²θ) = 1
2sin2θ(-cos2θ) = 1
2sin2θ(cos2θ) = -1
sin4θ= -1
4θ = sin⁻¹(-1)
4θ = 2nπ – π/2, n ϵ Z