Solving a sin/cos equation with identities

ChristaJoy

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Sep 23, 2012
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I need to solve this:
2sin^2(theta)=3(1-cos(-theta))

I understand that sin^2(theta)=1-cos^2(theta), but I don't know how that would do me any good to solve the problem. Im also stuck on this one:

cos(theta)-sin(-theta)=0

How would I go about solving this? Thank you all in advance :)
 
I need to solve this:
2sin^2(theta)=3(1-cos(-theta))

I understand that sin^2(theta)=1-cos^2(theta), but I don't know how that would do me any good to solve the problem. Im also stuck on this one:

cos(theta)-sin(-theta)=0

How would I go about solving this? Thank you all in advance :)

2sin^2(theta)=3(1-cos(-theta))

2sin2(Θ) = 3 * [1 - cos(-Θ)]

2[1 - cos2(Θ)] = 3 * [1 - cos(Θ)]

The above will give you a quadratic equation - which you should be able to solve.

for # 2

Hint:

sin[(2n+1)π + Θ] = -sin(Θ)
 
Hello, ChristaJoy!

\(\displaystyle 2\:\!\sin^2\theta \:=\:3[1-\cos(\text{-}\theta)]\)

Note that: \(\displaystyle \cos(\text{-}\theta) \,=\,\cos\theta\)

We have:
. . . . \(\displaystyle \begin{array}{ccc} 2\:\!\sin^2\theta \;=\;3(1-\cos\theta) \\ 2\:\!\sin^2\theta -3(1-\cos\theta) \;=\;0 \\ 2(1-\cos^2\theta) - 3(1-\cos\theta) \;=\;0 \\ 2(1-\cos\theta)(1+\cos\theta) - 3(1-\cos\theta) \;=\;0 \\ (1-\cos\theta)\big[2(1+\cos\theta) - 3\big] \;=\;0
\end{array}\)

Now solve: .\(\displaystyle \begin{Bmatrix}1-\cos\theta \;=\;0 \\ 2(1 + \cos\theta)-3 \;=\;0 \end{Bmatrix}\)



\(\displaystyle \cos\theta-\sin(\text{-}\theta)\:=\:0\)

Note that: \(\displaystyle \sin(\text{-}\theta) \:=\:\text{-}\sin\theta\)

We have:

. . . \(\displaystyle \begin{array}{cc}\cos\theta - (\text{-}\sin\theta) \;=\;0 \\ \cos\theta + \sin\theta \;=\;0 \\ \sin\theta \;=\;\text{-}\cos\theta \\ \dfrac{\sin\theta}{\cos\theta} \;=\;\text{-}1 \\ \tan\theta \;=\;\text{-}1 \end{array}\)

Got it?
 
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