Solve for [0, 360)
2sinxcosx + cosx =0
2sinxcosx = -cosx
2sinx = -cosx/cosx
sinx = -1/2
{210, 330)
Is this correct?
no , you only have half the answers
2sinxcosx + cosx =0
cosx(2sinx + 1) = 0
cosx = 0 or sinx = -1/2
if cosx = 0, x = 90° , 270°
if sinx = -1/2, x = 210°, 330°
(you only had the last case)
To solve the equation 2sin(x)cos(x) + cos(x) = 0 in the interval [0, 360), you have correctly simplified it to 2sin(x)cos(x) = -cos(x).
Now, we can simplify this equation further by dividing both sides by cos(x):
2sin(x) = -cos(x)/cos(x)
Since cos(x)/cos(x) equals 1, the equation becomes:
2sin(x) = -1
To find the solutions for sin(x) = -1, we need to recall the unit circle and its angles where sin(x) = -1. The unit circle represents all possible values of sine and cosine for angles from 0 to 360 degrees.
On the unit circle, the angles where sin(x) = -1 are 210 degrees and 330 degrees.
Therefore, the solutions to the equation 2sin(x)cos(x) + cos(x) = 0 in the interval [0, 360) are x = 210 degrees and x = 330 degrees.
So, your solution {210, 330) is correct!