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#1 2006-11-24 09:22:53

Fonzie
Guest

Trig functions?

Anyone know how to solve:

2sin^2x=3sinx cosx + 2cos^2x

Thanks

#2 2006-11-24 11:56:23

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

Re: Trig functions?

There may be an easier way this this, but you can do it this way:

2sin²x = 3sinxcosx + 2cos²x
Rearrange: 2(sin² - cos²x) = 3sinxcosx
Square both sides: 4(sin[sup]4[/sup]x - 2sin²xcos²x + cos[sup]4[/sup]x) = 9sin²xcos²x

Rearrange: 4(sin[sup]4[/sup]x + cos[sup]4[/sup]x) = 17sin²xcos²x

Now we can use the identity that sin²x + cos²x = 1.
Rearranging that gives cos²x = 1 - sin²x, which we can substitute into our equation.

4(sin[sup]4[/sup]x + (1 - sin²x)²) = 17sin²x(1 - sin²x)

Multiply out: 4(2sin[sup]4[/sup]x  - 2sin²x + 1) = 17(sin²x - sin[sup]4[/sup]x)

Rearrange: 25sin[sup]4[/sup]x - 25sin²x + 4 = 0

Now we can make a substitution of u = sin²x, and the equation becomes 25u² - 25u + 4 = 0, a quadratic equation.

Factorise: (5u - 4)(5u - 1) = 0.
Solve: u = sin²x = 1/5 or 4/5.

So then you have 4 solutions for sinx, ±1/√5 and ±2/√5.
You can then use the inverse sin function to find solutions for x, making sure you include any others that you need within your range.

For example, sin[sup]-1[/sup]1/√5 ≈ 26.6[sup]o[/sup], but you would also need to include the solution of 153.4[sup]o[/sup] if your range was from 0[sup]o[/sup] to 180[sup]o[/sup].

There may well be an easier way to get this, but I'm pretty sure that this way works.


Why did the vector cross the road?
It wanted to be normal.

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