2cosx+sinx+1=0

how do I get everything in one function?

2cosx+sinx+1=0

2cosx = -sinx-1
square both sides
4cos^2 x = sin^2 x + 2sinx +1
4(1-sin^2 x) = sin^2 x + 2sinx +1
4 - 4sin^2 x = sin^2 x + 2sinx +1
5sin^2 x + 2sinx -3 = 0
(sinx+1)(5sinx-3)=0
sinx=-1 or sinx = 3/5
x = 270 degrees or x = 36.9 degrees or 143.1 degrees.

but remember we squared, so all answers must be verified in the original equation

if x = 270
LS = 2cos 270 + sin270 + 1
= 0-1+1 = 0 = RS

if x= appr. 36.9
LS = 2cos36.9+sin36.9+1
= not the right side

so x = 270 degrees or 3pi/2 radians

To get everything in one function, you can rearrange the given equation by isolating all the trigonometric terms on one side and the constant term on the other side. Here's how:

1. Start with the given equation: 2cos(x) + sin(x) + 1 = 0.
2. Subtract 1 from both sides to isolate the trigonometric terms: 2cos(x) + sin(x) = -1.
3. Rearrange the terms: sin(x) = -2cos(x) - 1.
4. Now, to express everything on one side in terms of a single function, you can use the trigonometric identity: sin(x) = cos(π/2 - x).
5. Substitute this identity into the equation: cos(π/2 - x) = -2cos(x) - 1.
6. Simplify and rearrange the equation: cos(π/2 - x) + 2cos(x) = -1.

Now, you have everything in one function: f(x) = cos(π/2 - x) + 2cos(x) = -1.

Keep in mind that this is just one way to express the equation as a single function. Depending on the context or requirements, there might be alternative forms or simplifications possible.