Question

What values for θ (0 ≤ θ ≤ 2π) satisfy the equation?2 sin θ cos θ + root3 cos θ = 0?

Answer 1

`\bf 2sin(\theta )cos(\theta )+√(3)cos(\theta )=0\implies \stackrel{\textit{common factor}}{cos(\theta )[2sin(\theta )+√(3)]=0} \\\\[-0.35em] ~\dotfill\\\\ cos(\theta )=0\implies \theta =cos^(-1)(0)\implies \theta = \begin{cases} (\pi )/(2)\\\\ (3\pi )/(2) \end{cases} \\\\[-0.35em] ~\dotfill`

`\bf 2sin(\theta )+√(3)=\implies 2sin(\theta )=-√(3)\implies sin(\theta )=-\cfrac{√(3)}{2} \\\\\\ \theta =sin^(-1)\left( -\cfrac{√(3)}{2} \right)\implies \theta= \begin{cases} (4\pi )/(3)\\\\ (5\pi )/(3) \end{cases}`

Answer 2

Step-by-step answer:

Given equation:

2sin(theta)cos(theta) + sqrt(3)*cos(theta) = 0 ...........................(1)

Solve for theta for 0<=theta<=2pi.

Factor out cos(theta), we get

cos(theta) * ( 2sin(theta) + sqrt(3) ) = 0

By the zero product theorem, we can conclude

cos(theta) = 0 ...................................(2)

OR

2sin(theta) + sqrt(3) = 0 ................. (3)

Solving (2)

cos(theta) = 0 has solutions pi/2 or 3pi/2 from the cosing curve.

Solving (3)

2sin(theta) + sqrt(3) = 0 =>

sin(theta) = -sqrt(3)/2

which has solutions 4pi/3 or 5pi/3.

So the solutions to equation (1) are

S={pi/2, 4pi/3, 3pi/2, 5pi/3}