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I tried to solve this equation: $[\cos(2x)]^2-\sin(2x)=1$

I got different answer from the book. the answer in the book: $x=-45+180k$ , $x=90k$

Am I right? $x = -45+180k$ is equal to $x= 135+180k$ ? How can I check if it's the same? Or maybe I did a mistake? please help

\begin{align*} [\cos(2x)]^2-\sin(2x) & = 1\\ -\sin(2x)& = 1-\cos(2x)^2\\ -\sin(2x)& = \sin(2x)^2\\ -\sin(2x)-\sin(2x)^2 & = 0\\ \sin(2x)+\sin(2x)^2 & = 0\\ \sin(2x)[1+\sin(2x)] & = 0 \end{align*}

\begin{align*} \sin(2x) & =0 & 1 + \sin(2x) & = 0\\ 2x & = 180k & \sin(2x) & = -1\\ x & = 90k & 2x & = 270+360k\\ & & x & = 135+180k \end{align*}

N. F. Taussig
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Silas2033
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    Both answers are equivalent: Just notice that $$-45^{\circ}+180^{\circ}k=-45^{\circ}+180^{\circ}+180^{\circ}(k-1)=135^{\circ}+180^{\circ}(k-1)$$ And $k-1$ is an integer number if $k$ is. – Ángel Mario Gallegos Sep 29 '15 at 17:53
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    You can post mathematical notation using MathJax here, if you learn a little $\LaTeX$. I'd like to help get you started by editing your post, but cos2x^2 is ambiguous (needs some more parentheses to be clear). – hardmath Sep 29 '15 at 17:53
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    By the way, you've been here six hours and posted six very similar questions about trigonometric equations. Perhaps you should slow down and study the answers you've already gotten before posting more of these. – hardmath Sep 29 '15 at 17:56
  • @Mario G ,thank you for the note in your second comment. I edited the message. and thank you also for the help :) . hardmath , thank you for your editing. – Silas2033 Sep 29 '15 at 18:07
  • The nice formatting was done by @mathreadler and not me! – hardmath Sep 29 '15 at 18:12
  • Sorry I should not have edited it if I saw hardmath was on it. – mathreadler Sep 29 '15 at 18:13

1 Answers1

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$$\cos²(2x)-\sin(2x) = 1$$

$$-\sin(2x)= 1-\cos²(2x)$$

$$-\sin(2x)=\sin²(2x)$$

$$0= \sin²(2x)+\sin(2x)$$

$$\sin(2x)(\sin(2x)+1)=0$$

Then, you get two equations:

$$\sin(2x)=0$$

and

$$\sin(2x)+1=0$$

$$\sin(2x)=\sin (180)$$

and

$$\sin(2x)=\sin(270)$$

From the $$\sin(2x)=\sin(180)$$, you can get the solution:

$$2x=180+k.360$$ or $$x=90+k.180$$

and

$$2x= k.360$$ or $$x= k.180$$

From the $$\sin(2x)=\sin(270)$$, you can get the solution:

$$2x=270+k.360$$ or $$x=135+k.180$$

and

$$2x=(180-270)+k.360$$ or $$x= -45+k.180$$

So, the solution:

$x=k.180$

$x=90+k.180$

$x=135+k.180$

$x=-45+k.180$

John_dydx
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akusaja
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  • You have repeated the solution sketched by the OP with some extra flourishes, but I'm not sure this is as helpful in explaining the concern about possible difference between $-45 + k\cdot 180$ and $135 + k\cdot 180$ in degrees ($k$ an integer) as done earlier in the Comment by @MarioG – hardmath Sep 29 '15 at 18:10
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    Just some hints on typesetting, multiplication symbol when writing equations you can write \cdot also a backslash before sin or cos make them look less like variable names. – mathreadler Sep 29 '15 at 18:10
  • @hardmath , so actually, the solutions are only two? I think the OP only missed a little things in his/her solution, so it made confusion for OP. For mathreadler, thanks. I'll try to correct it in another occasion! – akusaja Sep 29 '15 at 18:13
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    look at the first line, $\cos$ should be in lower case, its cos not Cos @akusaja. –  Sep 29 '15 at 18:21