Find all the values of x between 0 and 2(pi) for which

sin x + (root 3)cos x = 0

Whenever you see sines and cosines and sqrt 3, draw a thirty, sixty ninety triangle.

the sin of the 60 degree angle is (sqrt 3)/2
the cosine of the 60 degree angle is 1/2
sqrt(3)/2 = (sqrt 3)(1/2)
Now the quadrants where the sine and cosine have opposite signs are quadrant 2 and quadrant four. So look in those two quadrants.

another not necessarily better approach would be ...

sinx = -√3cosx
tanx = -√3

from the same triangle as Damon's we would get a "basic" angle of 60º or pi/3 radians

but the tan is negative in II or IV, so
x = pi - pi/3 or 2pi - pi/3
or x = 2pi/3 or 5pi/3

To find all the values of x between 0 and 2π for which sin x + √3 cos x = 0, we can use the trigonometric identity for the sum of angles:

sin(a + b) = sin a cos b + cos a sin b

Comparing this identity to the given equation sin x + √3 cos x = 0, we can rewrite it as:

sin x + √3 cos x = sin(π/3) cos x + cos(π/3) sin x

Now we can apply the identity to rewrite the equation as:

sin(x + π/3) = 0

To find the values of x that satisfy this equation, we need to find the angles whose sine is zero. In other words, we need to find the angles x + π/3 where sin(x + π/3) = 0.

The sine of an angle is zero when the angle is a multiple of π. Therefore, x + π/3 = nπ, where n is an integer.

Solving for x, we have:

x = nπ - π/3

To find all the values of x between 0 and 2π, we need to substitute different integer values for n and check if they fall within the given range.

For example, when n = 0:
x = 0 - π/3 = -π/3 (outside the range)

When n = 1:
x = π - π/3 = 2π/3

Similarly, when n = 2:
x = 2π - π/3 = 5π/3 (outside the range)

We have found two values of x that satisfy the equation sin x + √3 cos x = 0 within the range of 0 to 2π: x = 2π/3 and x = 5π/3.