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**RachelSmith****Guest**

Hey there!

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!

**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

The identity you want here is cos² θ + sin² θ = 1.

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.

Why did the vector cross the road?

It wanted to be normal.

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**amonjezi****Member**- Registered: 2006-03-09
- Posts: 9

3cos^2x+3sinx=3

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)

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**amonjezi****Member**- Registered: 2006-03-09
- Posts: 9

thank you rashel

your question is very good

by

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