Question

Solve the equation
5sin (0) cos(0) + 4cos (0) = 0​

Answer 1

To solve the given trigonometric equation, we factor out the common term cos(θ), set each factor equal to zero, and solve for θ. There are two sets of solutions, one where cos(θ) = 0 and another where sin(θ) = -4/5.

Step-by-step explanation:

To solve the equation 5sin(θ)cos(θ) + 4cos(θ) = 0, we can factor out the common term, which is cos(θ). The equation then becomes:

cos(θ)(5sin(θ) + 4) = 0

For this equation to hold true, either cos(θ) = 0 or (5sin(θ) + 4) = 0 must be satisfied.

To find the solution where cos(θ) = 0, we use the fact that cosine is zero at θ = π/2 + kπ, where k is an integer. For the equation (5sin(θ) + 4) = 0, we solve for sin(θ) = -4/5. This latter solution is only valid if -1 ≤ sin(θ) ≤ 1, which is true in this case.

Answer 2

`\begin{aligned}\theta&=(\pi)/(2)+2\pi n\\\\\theta&= (3\pi)/(2)+2\pi n\\\\\theta&=-0.927+2\pi n\\\\\theta&=4.067+2 \pi n\end{aligned}`

Step-by-step explanation:

Given trigonometric equation:

`5\sin (\theta)\cos (\theta) + 4\cos (\theta) = 0`

Begin by factoring out the common term cos(θ):

`\cos (\theta)\left(5\sin (\theta) + 4\right) = 0`

Now, solve each part separately.

`\boxed{\begin{array}c\begin{aligned}\cos (\theta) & = 0\\\\\theta&=\arccos(0)\\\\\theta&=(\pi)/(2)+2\pi n, (3\pi)/(2)+2\pi n\\\\\\\\\end{aligned}&\begin{aligned}5\sin (\theta) + 4& = 0\\\\\sin (\theta) &=-(4)/(5)\\\\\theta &=\arcsin\left(-(4)/(5)\right)\\\\\theta &=-0.927+2\pi n, 4.067+2 \pi n\end{aligned}\end{array}}`

Therefore, the solutions to the given equation are:

`\boxed{\begin{aligned}\theta&=(\pi)/(2)+2\pi n\\\\\theta&= (3\pi)/(2)+2\pi n\\\\\theta&=-0.927+2\pi n\\\\\theta&=4.067+2 \pi n\end{aligned}}`

Please note that the solutions are in radians, and are rounded to 3 decimal places where appropriate.