cos2theta = 3sintheta + 2
First, we make the problem easier by setting: Why? because this puts us in a linear space; linear is easy. sin theta = x cos theta = y which gives 3x-4y=2 or y=3/4x-1/2 (eq. 1) if we remember our trig identities now we also know sin^2 + cos^2 =1 so x^2+y^2=1 (eq. 2) so now that we have two independent equations for x and y, we can solve for them by reduction replace y in eq 2 with eq 1 and solve for x. (this is a nasty one.) x^2 + (3/4x -1/2)^2 = 1 x^2 + 9/16x^2 - 6/8x +1/4 = 1 25/16x^2-6/8x-3/4=0 25x^2-12x-12=0 use the quadratic equation to yield x=-0.493 OR x=0.973 since x= sin theta we plug in our two x's (one at a time) to reveal: theta = 76.7 degrees or -29.6 degrees why two answers? since sin theta, cos theta sweep a circle, a linear relationship can have 0, 1, or 2 solutions.
I hope this helps hun ^^^
hun, huh? Are you from Baltimore of the south? (Just wondering)
*or the
Larry i put something above it will explain xD
I know the answer, I was asking you personally :)
\(\cos(2\theta) = 3\sin(\theta) + 2\) \(1 - 2\sin^{2}(\theta) = 3\sin(\theta) + 2\) \(2\sin^{2}(\theta) + 3\sin(\theta) + 1 = 0\) \((2\sin(\theta) + 1)(\sin(\theta) + 1) = 0\) \(\sin(\theta) = -1/2\;or\;\sin(\theta) = -1\) You're almost done. Of course, if you meant \(\cos^{2}(\theta)\), then that's a horse of a different color.
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