your question is very good

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3(1-sin^2x)-sinx-2=0

3-3sin^2x-sinx-2=0

3sin^2x+sinx-1=0

if sinx=y

3y^2+y-1=0

this equaition very easy (0 degree to 360 degree)]]>

Using this, you can convert the cos² x's into sin² x's. Then you've got a nice, easy quadratic equation which will let you find sin x. Then just take the inverse of that to find x.

]]>I would appreciate it so much if someone could help me with these! I need the method and the answer, because the way I'm doing it seems to be all wrong.. I am not even sure I know what I'm doing.. So I need someone to explain it to me.. THANK YOU so much!

Use identities to obtain approximate solutions to these equations giving all answers to the nearest degree:

1) 3cos^2 x - sin x - 2 = 0, 0 _< x _< 360 degrees

2) 2cos^2x + 3 sin x = 3, 0 _< x _< 180 degrees

thank you so much!

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