I'm having difficulty solving this equation and I think I keep going about it wrong.

Solve the equation cos2x + 2sinx = sin²x in the interval 0≤ x < 360.

Thanks in advance ^_^

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- May 8th 2011, 03:00 AMHighlyFlammableEquation Solving
I'm having difficulty solving this equation and I think I keep going about it wrong.

Solve the equation cos2x + 2sinx = sin²x in the interval 0≤ x < 360.

Thanks in advance ^_^ - May 8th 2011, 03:25 AMHappyJoe
Here's an idea:

Use cos(2x) = cos^2(x) - sin^2(x), and use cos^2(x) = 1 - sin^2(x) to rewrite the equation into something that involves only sin(x). You should get a quadratic in sin(x), which you can solve with the usual methods. - May 8th 2011, 03:34 AMHighlyFlammable
Thank you ^_^ I didn't realise I could use cos^2(x) = 1 - sin^2(x). I'll have to remember to bear that in mind :)

- May 8th 2011, 05:58 AMHighlyFlammable
Hi,

I have another question I'm unsure of.

If f(x) = 2sin(3x - π/2) + 5, what is the range of values of f(x)?

Thank you again :) - May 8th 2011, 06:09 AMe^(i*pi)
You can tell (either by knowing or looking at the graph) that: $\displaystyle -1 \leq \sin(\theta) \leq 1$.

The minimum value of f(x) is $\displaystyle f(x) = 2 \cdot -1 + 5$. Work out the maximum value and that will give you the range of f(x)