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Math Help - Solve the equation

  1. #1
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    Solve the equation

    Can someone help me please. I don't know which way to go with this.

    Solve the equation

    sin2x - 1 = cos 2x

    giving answers for values of x between 0 and 360 degrees.
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  2. #2
    MHF Contributor Amer's Avatar
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    use this for

    sin2(x)=2sin(x)cos(x)

    and

    cos2(x)=2cos^2(x)-1

    sub these and find the solution
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  3. #3
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    Hello, jo74!

    Solve the equation: . <br />
\sin2x - 1 \:= \:\cos 2x\;\;\text{ for }0^o < x < 360^o
    Use the identities: . \begin{array}{ccc}\sin2\theta &=& 2\sin\theta\cos\theta \\ \cos2\theta &=& 2\cos^2\!\theta - 1\end{array}

    And we have: . 2\sin x\cos x - 1 \;=\;2\cos^2\!x - 1 \quad\Rightarrow\quad 2\sin x\cos x - 2\cos^2\!x \;=\;0


    Factor: . 2\cos x(\sin x - \cos x) \:=\:0


    And we have two equations to solve:

    . . 2\cos x \:=\:0 \quad\Rightarrow\quad\boxed{x \:=\:90^o,\:270^o}

    . . \sin x - \cos x \:=\:0 \quad\Rightarrow\quad \sin x \:=\:\cos x
    . . . . Divide by \cos x\!:\;\;\frac{\sin x}{\cos x} \:=\:1 \quad\Rightarrow\quad \tan x \:=\:1 \quad\Rightarrow\quad\boxed{x \:=\:45^o,\:225^o}


    Edit: Ah . . . Amer already explained the game plan . . .
    .
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by jo74 View Post
    Can someone help me please. I don't know which way to go with this.

    Solve the equation

    sin2x - 1 = cos 2x

    giving answers for values of x between 0 and 360 degrees.
    Here's a proof of the double angle formula for the sine and cosine functions so that you can better understand them.





    sin 2α = sin (α + α)
    sin 2α = sin α cos α + cos α sin α
    sin 2α = 2 sin α cos α

    So, there's two ways to substitute here. The question you have to ask yourself is which one will be more efficacious.
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