Question

Solve the equation: √3 sin θ - cos θ = √2

Answer 1

Answer:

$\theta=\frac{5\pi }{12}$

or

$\theta=\frac{-\pi }{12}$

Step-by-step explanation:

to solve the given equation

√3 sin θ - cos θ = √2

divide both side by 2

$\frac{\sqrt{3} }{2}\:sin\:\theta-\frac{1}{2}cos\:\theta=\frac{\sqrt{2}}{2}\\\\\frac{\sqrt{3} }{2}\:sin\:\theta-\frac{1}{2}cos\:\theta=\frac{1}{\sqrt{2}}$

As we know that

$cos\:\frac{\pi }{6} =\frac{\sqrt{3} }{2} \\\\sin\:\frac{\pi }{6} =\frac{1}{2}$

$cos\frac{\pi }{6} \:sin\:\theta-sin\frac{\pi }{6} cos\:\theta=\frac{1}{\sqrt{2}}$

as we know that

$sin(A-B)=sinAcosB-cosAsinB$

$cos\frac{\pi }{6} \:sin\:\theta-sin\frac{\pi }{6} cos\:\theta=\frac{1}{\sqrt{2}}\\\\sin(\frac{\pi }{6}-\theta)= \frac{1}{\sqrt{2}}\\\\(\frac{\pi }{6}-\theta)=sin^{-1}(\frac{1}{\sqrt{2}})\\\\(\frac{\pi }{6}-\theta)=\frac{\pi }{4}\\ \\\theta=\frac{\pi }{6}-\frac{\pi }{4}\\\\\theta=\frac{-\pi }{12}$

or

$cos\frac{\pi }{6} \:sin\:\theta-sin\frac{\pi }{6} cos\:\theta=\frac{1}{\sqrt{2}}\\\\sin(\theta-\frac{\pi }{6})= \frac{1}{\sqrt{2}}\\\\(\theta-\frac{\pi }{6})=sin^{-1}(\frac{1}{\sqrt{2}})\\\\(\theta-\frac{\pi }{6})=\frac{\pi }{4}\\ \\\theta=\frac{\pi }{6}+\frac{\pi }{4}\\\\\theta=\frac{5\pi }{12}$

Answer 2

Answer:

Step-by-step explanation:

Formula used:

$The\:solution\:of\:\cos\theta=cos\alpha\:is\:\theta=2n\pi{+_-}\alpha$

$\sqrt{3}sin\theta-cos\theta=\sqrt{2}$

$cos\theta-\sqrt{3}sin\theta=-\sqrt{2}$

divide both sides by 2

$\frac{1}{2}cos\theta-\frac{\sqrt{3}}{2}sin\theta=-\frac{\sqrt{2}}{2}$

$\frac{1}{2}cos\theta-\frac{\sqrt{3}}{2}sin\theta=-\frac{1}{\sqrt{2}}$

$cos(\theta-\frac{\pi}{3})=cos\frac{3\pi}{4}$

$Therefore$

$\theta-\frac{\pi}{3}=2n\pi+\frac{3\pi}{4}\\or\\\theta-\frac{\pi}{3}=2n\pi-\frac{3\pi}{4}$

$\theta=2n\pi+\frac{3\pi}{4}+\frac{\pi}{3}\\or\\\theta=2n\pi-\frac{3\pi}{4}+\frac{\pi}{3}$

where n is an integer