Question

Solve the trigonometric equation over the interval [0, 2π)
$\large{ {cos}^{2} \theta + \sqrt{3} sin \theta cos \theta = 1}$
Show your work, thanks!​

Answer 1

Answer:

θ = 0° , 60°

Step-by-step explanation:

$\cos^2\theta + \sqrt{3}\sin \theta \cos \theta = 1$

• Substract both the sides from 1.

$=> \cos^2\theta + \sqrt{3}\sin \theta \cos \theta - 1 = 1 - 1$

$=> \cos^2\theta + \sqrt{3}\sin \theta \cos \theta - 1 = 0$

• Rearrange the terms in L.H.S.

$=> (\cos^2 \theta - 1 )+ \sqrt{3} \sin \theta \cos \theta = 0$

• Replace the terms in brackets with '-sin²θ' by using the identity (sin²θ + cos²θ = 1).

$=> -\sin^2 \theta + \sqrt{3} \sin \theta \cos \theta = 0$

• Take 'sinθ' common from the whole expression in L.H.S.

$=> \sin \theta(- \sin \theta + \sqrt{3}\cos \theta) = 0$

Now solve each part separately :-

1) $\sin \theta = 0$

$=> \theta = \sin^{-1}0 = 0$

2) $-\sin \theta + \sqrt{3} \cos \theta = 0$

• Divide both the sides by 'cosθ'.

$=> \frac{-\sin \theta + \sqrt{3}\cos \theta }{\cos \theta} = \frac{0}{\cos \theta}$

$=> \frac{-sin \theta}{\cos \theta} + \frac{\sqrt{3} \cos \theta}{\cos \theta} = 0$

$=> -\tan \theta + \sqrt{3} = 0$

• Add both the sides with 'tanθ'.

$=> -\tan \theta + \sqrt{3} + \tan \theta = 0 + \tan \theta$

$=> \tan \theta = \sqrt{3}$

$=> \theta = \tan^{-1}\sqrt{3} = 60$

∴ θ = 0° , 60°

Answer 2

Answer:

Solution given:

$\large{ {cos}^{2} \theta + \sqrt{3} sin \theta cos \theta = 1}$

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

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

$Sin\theta(sin\theta -\sqrt{3}cos\theta)$

Either

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

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

$Tan \theta =\sqrt{3}$

$Tan \theta =Tan 60,Tan(180+60),Tan (360+60)$

$\theta =60°,240°,420°$

in terms of π is

$\theta =⅓π,\frac{4}{3}π,$

Or

$Sin\theta=1$

$Sin\theta=Sin0° ,Sin180°$

In terms of π is

$\theta=0π,π$

so

$\theta=0π,⅓π,π,\frac{4}{3}π,$