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I am solving $\sin 2x = \sin x +\cos x $ for $0\le x \le 360$

$$\sin 2x = \sin x +\cos x$$ $$2 \sin x \cos x =\sin x + \cos x$$ $$(\cos x + \sin x) ^{2} - (\sin x)^{2} - (\cos x)^{2} =(\cos x + \sin x)^{2} - 1=\sin x + \cos x$$ Let $\cos x + \sin x = y$ $$y^{2} - 1= y$$ After solving the quadratic gets $1.618$(I think this is not accepted) and $-0.618$ $$y=\cos x + \sin x =\sin 2x = -0.618$$ $$2x=218.17,321.83,578.17,681.87$$ $$x=109.09,160.92,289.09,340.34$$ But the problem is $109.09$ and $340.34$ is not the solution. I didn't purposely square anything to produce extra solution. The two extra solution satisfy $\sin 2x=-0.618$ but not $\cos x + \sin x=-0.618$.Is there any mistake in my calculations, or is there anyplace I introduced accidentally extra solution? Thanks.

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  • $\begingroup$ I don't get how you get from line (2) to line (3) : there should be extra $\sin x + \cos x$ at the RHS! $\endgroup$ – Wyllich Sep 18 '17 at 11:48
  • $\begingroup$ @Wyllich I just show the left hand side of the equation in line 3. $\endgroup$ – lyk Sep 18 '17 at 11:52
  • $\begingroup$ @lyk Sorry, but I still cannot interpret what you have done from line (2) to line (3). Keep in mind that you must be careful with trigonometry, you cannot simply 'cancel' from both sides. $\endgroup$ – Landuros Sep 18 '17 at 11:55
  • $\begingroup$ @Landuros Is that clearer now? $\endgroup$ – lyk Sep 18 '17 at 12:02
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    $\begingroup$ @Landuros the $\sin x +\cos x $ is the right hand side of the equation ,i didnt make the left hand part to become the right hand part of the original equation. $\endgroup$ – lyk Sep 18 '17 at 12:21
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Consider these three equations:

\begin{align} \sin 2x = \sin x + \cos x \tag{1} \\ (\sin x + \cos x)^2 - 1 = \sin x + \cos x \tag{2} \\ (\sin 2x)^2 -1 = \sin 2x \tag{3} \end{align}

You were correct in determining that (1) and (2) are equivalent. Also, (1) does imply (3).

However, what you failed to notice is that (3) does not imply (1), so while all solutions to

\begin{align} y^2 - 1 &= y, \\ y &= \sin x + \cos x \end{align} are solutions to (1), the same isn't true for \begin{align} y^2 - 1 &= y, \\ y &= \sin 2x. \end{align}


To illustarte this, here is a example using only basic algebra and not trigonometry, which might look stupid, but is logically analogous to your soliution:

Problem: Solve $$ x^2 - x = x - 1. \tag{1} $$

First, we add $1-x$ to both sides to see that this is equivalent to $$ (x-1)^2 = 0. \tag{2} $$ Then, we denote $y=x-1$ to solve $y^2 = 0$ it and get $y=0$. So far so good.

Finally, we use $y = x^2 - x$ (This is a stupid way to do this but it illustrates the error.), and solve $$ x^2 - x = 0, $$ and find $x=1$ and $x=0$ as the solutions.

Where did we go wrong?

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    $\begingroup$ Why (3) does not imply (1)? $\endgroup$ – lyk Sep 18 '17 at 12:30
  • $\begingroup$ you square.............. $\endgroup$ – Aryadeva Sep 18 '17 at 12:33
  • $\begingroup$ @lyk Because it has solutions that are not solutions to (1), as you noticed. I'll try to come up with a simpler example of similar error without trigonometry. $\endgroup$ – JiK Sep 18 '17 at 12:34
  • $\begingroup$ @JiK How could I know $y =\sin 2x $ had more solutions than $y=\sin x + \cos x$ without drawing graph? $\endgroup$ – lyk Sep 18 '17 at 13:03
  • $\begingroup$ @lyk You don't need to know it a priori. You just should be careful in your logical process to see which equations actually imply which equations. $\endgroup$ – JiK Sep 18 '17 at 13:05
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Sometimes a graph helps in understanding problems like this. Here are the graphs of the curves $y = \sin(2x^\circ)$ and $y = \sin x^\circ + \cos x^\circ$ as well as the line $y=-0.618$ for $0 \leq x \leq 360,$ plotted by https://www.desmos.com/calculator:

enter image description here

As you can see, the graphs of $y = \sin(2x^\circ)$ and $y = \sin x^\circ + \cos x^\circ$ cross only twice in this region, but $y = \sin(2x^\circ)$ crosses the line $y=-0.618$ four times. That's where your two incorrect "solutions" are coming from.

Since you found it difficult to solve $\sin x^\circ + \cos x^\circ = -0.618,$ a way to avoid wrong answers is to solve $\sin(2x^\circ) = -0.618$ first and then check each of the solutions of $\sin(2x^\circ) = -0.618$ to see if that value of $x$ also makes it true that $\sin(2x^\circ) \approx \sin x^\circ + \cos x^\circ.$ You should easily be able to confirm that $\sin(2x^\circ) \approx \sin x^\circ + \cos x^\circ$ if $x=160.92$ or $x=289.09$, but that $\sin(2x^\circ) \not\approx \sin x^\circ + \cos x^\circ$ if $x=109.09$ or $x=340.34.$

If you actually want to try that on-line graphing tool yourself, you have to convert from degrees to radians inside each of the functions, since Desmos expects the sine and cosine to be functions of radians rather than degrees. The formulas I used were y=sin(2x*\pi/180) and y=sin(x*\pi/180)+cos(x*\pi/180).

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  • $\begingroup$ I actually plug in the answer in $y=\sin 2x$ to $y=\sin x +\cos x $ and know that the two solution are extra. After reading JiK's answer I know that my reasoning is wrong, and after reading your answer, this confirms my thought that $-0.618=\sin 2x$ has more solution than $-0.618=\sin x +\cos x$,so this produce extra solutions,as I make the error of $-0.618=\sin x + \cos x=\sin 2x$ to get the two extra solution of $-0.618=\sin 2x$ $\endgroup$ – lyk Sep 19 '17 at 9:59
  • $\begingroup$ Just to be clear, there's nothing wrong with using $\sin 2x = -0.618$ and getting four angles from it. That is a completely legitimate technique to use while working on this problem. It only becomes a mistake if you forget to check that $\sin x + \cos x = -0.618$ afterwards. $\endgroup$ – David K Sep 19 '17 at 11:39
  • $\begingroup$ I actually had check $\sin x + \cos x=-0.618$ so that I know the two solutions are extra, I am here to ask why I get the extra solution. I think my mistake is that I don't know $\sin 2x=-0.618$ and $\sin x + \cos x=-0.618$ have different number of solutions,(I don't use graph).And also why JiK say that this is an error? I am convinced by him that I should avoid using this way to solve equation to avoid getting extra solution(left hand side and right hand side of equation have different number of solution) $\endgroup$ – lyk Sep 19 '17 at 13:57
  • $\begingroup$ When I read the question, I mistakenly thought that you found out there were only two solutions from the answer key rather than from actually checking both equations. Perhaps JiK also had that thought. Of course it is fine if you can avoid generating the false "solutions" in the first place, but the important thing is that in the end you list all the correct solution values and do not list any incorrect ones. $\endgroup$ – David K Sep 19 '17 at 18:27
  • $\begingroup$ Actually there are no answer key at all..... $\endgroup$ – lyk Sep 19 '17 at 22:30
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HINT

Please follow the link:

https://www.wolframalpha.com/input/?i=sin(2x)%3Dsin(x)%2Bcos(x)

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  • $\begingroup$ I saw the equation have 2 real solutions in the link, but why I found extra solutions? $\endgroup$ – lyk Sep 18 '17 at 11:56
  • $\begingroup$ @lyk Because of the square $((cos(x)+sin(x))^2.... $ in $(cosx+sinx)2−(sinx)2−(cosx)2=(cosx+sinx)2−1=sinx+cosx$ $\endgroup$ – Aryadeva Sep 18 '17 at 12:24
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    $\begingroup$ @Isham i didnt square it, it just the identity of $2ab=a^{2} + b^{2} +2ab-a^{2} - b^{2} $ $\endgroup$ – lyk Sep 18 '17 at 12:34
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You have to do it in a systematic way. Once you substitute $y=\cos x+\sin x$ and have found the value for $y=-0.618$, go back to the substitution and solve $y=\cos x+\sin x=-0.618$. When solving $\sin 2x=-0.618$ you cannot just forget that $\sin 2x=\cos x+\sin x$.

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  • $\begingroup$ I cant find a way to $y=\cos x+\sin x=-0.618$.So I am forced to solve $y=\cos x+\sin x=\sin 2x=-0.618$. $\endgroup$ – lyk Sep 18 '17 at 12:05
  • $\begingroup$ @lyk Just solve $\cos x+\sin x=-0.618$. Use the rewriting $\cos x+\sin x=\sqrt{2}\sin(x+\pi/4)$. $\endgroup$ – A.Γ. Sep 18 '17 at 12:10
  • $\begingroup$ Unfortunately, the rewriting is out of my school syllabus.This actually is a question asked by my teacher to solve it at home.Is there a more simple form to solve?But quite frankly,I can solve this using the rewrited form,but how can I show to my teacher? $\endgroup$ – lyk Sep 18 '17 at 12:16
  • $\begingroup$ @lyk It is a simple rewriting and used to be a part of school education. All you need is just the formula for $\sin(x+y)$. $\endgroup$ – A.Γ. Sep 18 '17 at 12:23
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    $\begingroup$ @Isham So isnt the extra solution is 1.618 because of y^2?I still believe mymistake is using y=sin2 x $\endgroup$ – lyk Sep 18 '17 at 15:41

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