Question

What values for θ (0 ≤ θ ≤ 2π) satisfy the equation?2 sinθ cos θ + sqr rt 2 cos θ = 0

Answer 1

ANSWER

`(\pi)/(2), (3\pi)/(2), (5\pi)/(4), (7\pi)/(4)`

Step-by-step explanation

Given;

`2\sin \left(θ\right)\cos \left(θ\right)+√(2)\cos \left(θ\right)=0`

Factorise cos ;

`\cos \left(θ\right)\left(2\sin \left(θ\right)+√(2)\right)=0`

Solving separately, we have;

`\begin{gathered} \cos\left(θ\right)=0 \\ or \\ 2\sin \left(θ\right)+√(2)=0 \\ \end{gathered}`
`cos(\theta)=0\text{ }`

Cos theta will only be zero when angle is;

`(\pi)/(2),(3\pi)/(2)`

SOolving for;

`\:2\sin \left(θ\right)+√(2)=0`

The solution is;

`\begin{gathered} \begin{equation*} 2\sin\left(θ\right)+√(2)=0 \end{equation*} \\ =θ=(5\pi)/(4),(7\pi)/(4) \end{gathered}`

Hence, combining the two solutions , we have;

`(\pi)/(2),(3\pi)/(2),(5\pi)/(4),(7\pi)/(4)`

ANSWER