sin2x+cosx=0 , [-180,180)
= 2sinxcosx+cosx=0
= cosx(2sinx+1)=0
cosx=0
x1=cos^-1(0)
x1=90
x2=360-90
x2=270
270 doesn't fit in [-180,180) what do I do? Or maybe I did something wrong.
sinx=1/2
x1=sin^-1(1/2)
x1=30
x2=180-30
x2=150
is this correct?
Please and Thank you
but your domain is from -180 to 180 , so even though 270 will work in the equation , it is beyond your domain.
But, by looking at the cosine graph , you will see that it also has an x-intercept at x = -90°
so for your first one,
x = -90, 0, 90
the second part for sinx = 1/2
your answers of 30° or 150° are correct for the given domain.
Yes, you have correctly solved the equation sin(2x) + cos(x) = 0.
First, you expanded sin(2x) to get 2sin(x)cos(x) + cos(x) = 0.
Next, you factored out cos(x) from the equation, giving cos(x)(2sin(x) + 1) = 0.
Then, you set each factor equal to zero, solving for cos(x) and 2sin(x) + 1 individually.
For cos(x) = 0:
You used the inverse cosine function (cos^(-1)) to find the angle whose cosine is 0. This gives you x1 = 90 degrees.
You also correctly used the fact that cosine is an even function, so you found x2 = 360 - 90 = 270 degrees. However, this value does not fall within the range of [-180, 180).
Since the given range is [-180, 180), x2 = 270 degrees does not satisfy the condition. Therefore, you can't include x2 in the solution for this particular problem.
Now for 2sin(x) + 1 = 0:
You set 2sin(x) equal to -1 and solved for sin(x) using the inverse sine function (sin^(-1)). This gives you x1 = 30 degrees.
Again, you used the fact that sine is an odd function, so x2 = 180 - 30 = 150 degrees.
So, the correct solution within the given range is x = {90, 30} degrees.
Well done!