Find mistake in solving $\sin 2x=\sin x + \cos x$

Question

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.

Answer 1

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}

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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?

Answer 2

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:

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).

Answer 3

HINT

Please follow the link:

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

Answer 4

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 $\sin 2x=\cos x+\sin x$.