Getting ready for trig semester finals. Can't figure out how they are simplifying these equations. (these are the review not actual test)

tanx-sqr3=2tanx

and
2sin^2+sinx=0

tanx-sqr3=2tanx

tanx = -√3
tanx is negative , so x must be in quads II or IV
I know tan 60° = √3
so x = 180-60= 120°
or
x = 360-60 = 300°

in radians that would be 2π/3 or 5π/3

2sin^2 x + sinx = 0
sinx(2sinx + 1) = 0
sinx = 0 or sinx = -1/2
sinx = 0 ----> x = 0, 180° or 360°
or sinx = -1/2
since sinx is negative, x must be in III or IV
I know sin 30 = +1/2
x = 180+30 = 210 or x = 360-30 = 330°

so x = 0,180, 360,210 , or 330

I get the second one, but the first I don't. What identity are you using for the first to get from line one to two?

just rearranging terms, collecting tan's on the left, numbers on the right. Pretty mysterious, eh?

To simplify the first equation, let's start by merging the terms involving tan(x) on one side and the constant term on the other side.

1. tan(x) - √3 = 2tan(x)

First, let's bring all the tan(x) terms to one side of the equation:

tan(x) - 2tan(x) = √3

Next, combine the tan(x) terms:

-tan(x) = √3

Now, to solve for tan(x), divide both sides by -1:

tan(x) = -√3

The solution, therefore, is tan(x) = -√3.

Now let's move on to the second equation:

2sin^2(x) + sin(x) = 0

To simplify this equation, we can factor out sin(x):

sin(x)(2sin(x) + 1) = 0

From this factored form, we can see that the equation will be true if either sin(x) = 0 or 2sin(x) + 1 = 0.

1. sin(x) = 0

This equation implies that x could be any angle for which sin(x) = 0, such as x = 0, π, 2π, etc.

2. 2sin(x) + 1 = 0

Subtracting 1 from both sides:

2sin(x) = -1

Dividing both sides by 2 to isolate sin(x):

sin(x) = -1/2

This equation implies that x could be any angle for which sin(x) = -1/2. Using special right triangles or the unit circle, we can find two angles that satisfy this condition: x = 7π/6 and x = 11π/6 (among others).

Therefore, the solutions for this equation are x = 0, π, 2π, 7π/6, and 11π/6.