Posted by Anonymous on Wednesday, November 21, 2007 at 6:29pm.
3sinx = 2cos^2(x)
I believe you've treated this as cos(2x) and expanded it as a double-angle identity. However, you can't do this because the cosine function is being squared. The 2 is not being multiplied by the x.
I'm working on it...
Start by recalling the most important identity. My math teacher calls this "the #1 Identity."
sin^2(x) + cos^2(x) = 1
We want to simplify our trig equation by writing everything in terms of sine. Let's solve the #1 Identity for cos^2(x) because we have that in our trig equation.
cos^2(x) = 1 - sin^2(x)
Plug that into the trig equation, and see what you can get from there. Let me know if you get stuck along the way.
wow I forgot about that identity, thank you!
But, I'm still stuck...
I get
cos^2x + sin^2x = 1
I can't use any compound identities for this...I don't know how to isolate for either cos or sin...
The only identity you need it the #1 Identity.
Plugging in 1 - sin^2(x) for cos^2(x), you should get...
3sinx = 2(1 - sin^2(x))
Use the distributive property, collect everything on the left, and then factor.
The only identity you need is the #1 Identity.
Plugging in 1 - sin^2(x) for cos^2(x), you should get...
3sinx = 2(1 - sin^2(x))
Use the distributive property, collect everything on the left, and then factor.
thanks, I solved it :)
Glad to be of service. :)
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