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

Question

Solve the trigonometric equation over the interval [0, 2π)


`\large{ {cos}^(2) \theta + √(3) sin \theta cos \theta = 1}`
Show your work, thanks!​

Answer 1

θ = 0° , 60°

Explanation:

`\cos^2\theta + √(3)\sin \theta \cos \theta = 1`

• Substract both the sides from 1.

`=> \cos^2\theta + √(3)\sin \theta \cos \theta - 1 = 1 - 1`

`=> \cos^2\theta + √(3)\sin \theta \cos \theta - 1 = 0`

• Rearrange the terms in L.H.S.

`=> (\cos^2 \theta - 1 )+ √(3) \sin \theta \cos \theta = 0`

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

`=> -\sin^2 \theta + √(3) \sin \theta \cos \theta = 0`

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

`=> \sin \theta(- \sin \theta + √(3)\cos \theta) = 0`

Now solve each part separately :-

1)
`\sin \theta = 0`

`=> \theta = \sin^(-1)0 = 0`

2)
`-\sin \theta + √(3) \cos \theta = 0`

• Divide both the sides by 'cosθ'.

`=> (-\sin \theta + √(3)\cos \theta )/(\cos \theta) = (0)/(\cos \theta)`

`=> (-sin \theta)/(\cos \theta) + (√(3) \cos \theta)/(\cos \theta) = 0`

`=> -\tan \theta + √(3) = 0`

• Add both the sides with 'tanθ'.

`=> -\tan \theta + √(3) + \tan \theta = 0 + \tan \theta`

`=> \tan \theta = √(3)`

`=> \theta = \tan^(-1)√(3) = 60`

∴ θ = 0° , 60°

Answer 2

Solution given:

`\large{ {cos}^(2) \theta + √(3) sin \theta cos \theta = 1}`

`1-sin^(2) \theta+√(3) sin \theta cos \theta = 1`

`1-1 +sin^(2)\theta=√(3) sin \theta cos \theta = 1`

`Sin\theta(sin\theta -√(3)cos\theta)`

Either

`sin\theta=√(3) cos \theta`

`√(sin\theta){cos \theta}=√(3)`

`Tan \theta =√(3)`

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

`\theta =60°,240°,420°`

in terms of π is

`\theta =⅓π,(4)/(3)π,`

Or

`Sin\theta=1`

`Sin\theta=Sin0° ,Sin180°`

In terms of π is

`\theta=0π,π`

so

`\theta=0π,⅓π,π,(4)/(3)π,`