1. ## Trigonometry

1)
2cosx-sinx=3/2cosx+sinx
solve all possible ans of x.

2) I got cosx=-1 and 0.4
i am suppose to find all possible ans of x.
i got 66.4 and 293.6 for the 0.4 already.
but how to slove for the -1? the reference angle is 0 and since cos is negative, it should be in the 2nd and 3rd quarant. and thus i got 90 and 180. however the ans is only 180. why?

2. #1

Do you mean

$2 \cos(x) - \sin(x) = \frac{3}{2} \cos(x) + \sin(x)$

or

$2 \cos(x) - \sin(x) = \frac{3}{2 \cos(x) + \sin(x)}$

or something else?

#2

cos(90º) = 0

Are you to find ALL solutions? Perhaps the Domain is restricted in some way?

Check your definitions. Are the coordinate axes technically IN the various quadrants?

3. i mean the second option

4. Originally Posted by helloying
1)
2cosx-sinx=3/2cosx+sinx
solve all possible ans of x.

2) I got cosx=-1 and 0.4
i am suppose to find all possible ans of x.
i got 66.4 and 293.6 for the 0.4 already.
but how to slove for the -1? the reference angle is 0 and since cos is negative, it should be in the 2nd and 3rd quarant. and thus i got 90 and 180. however the ans is only 180. why?
Because cos(180) = -1 and cos(90) = 0. See this site and look just beneath the applet that graphs the cosine function (though this works just as well). There is a graph of y = cos(x) drawn in the "Did you notice" section.

-Dan

5. Originally Posted by helloying
i mean the second option
When there is a question, there is also a wonderful opportunity to provide some information about your efforts and your thoughts about solvign the problem.

I offer another opportunity for you to show your work and share your thoughts.

6. ## help me please! Trigonometry

12 cos x - 5 sin x = R cos (x + 22.6)

solve x in range 0 to 360 degrees
(12 cos x- 5 sin x)^2

Also, find the max and min value of eqn.

Kindly explain it clearly because I actually got an explanation from somewhere but it is not clear. Thank you greatly for your help.God bless you.

7. ## sorry post in wrong place

i wanted to post it in new thread but accidentally post it here. i have posted it in another new thread already.