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$$\sin 2x + \cos 2x =-1$$

How would I go about solving this equation? Some hint on how to start, so I can try to figure it out on my own. Thank you

8 Answers 8

6

$$\sin 2x+\cos 2x=-1$$ $$2\sin x\cos x+\cos^2x-\sin^2x+\sin^2x+\cos^2x=0$$ $$2\sin x\cos x+2\cos^2x=0$$ $$\cos x(\sin x+\cos x)=0$$ $$\cos x=0\Rightarrow x=\frac{\pi}{2}+k\pi,\tan x=-1\Rightarrow x=-\frac{\pi}{4}+k\pi$$

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    Yeah, that is what I got, and I got cosx=0 -> x = k*pi But how did you get tanx from sinx+cosx=0? Or am I missing something here – 2017-02-09
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    $\sin x+\cos x=0\Rightarrow \frac{\sin x}{\cos x}+1=0\Rightarrow\tan x=-1$ – 2017-02-09
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    yeah got it right now. Thank you for help. – 2017-02-09
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    you are welcomme and If you think that my answer is helpful you by right clicking can uppvote it and mark as accepted – 2017-02-09
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    For $\tan x$ the periodic term is $k\pi,$ not $k\pi/2.$ – 2017-02-09
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    yes, I see my mistake thank you for that. – 2017-02-09
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    $\cos x = 0$ does not mean $x=k\pi$, either. That would be sinus. – 2017-02-09
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Hint:

Write

$$\sin 2x=2\sin x\cos x$$ $$\cos 2x=2\cos^2x-1$$

and simplify.

2

If we consider $$A\sin 2x + B\cos 2x$$ could be of the form $$r \cos y \sin 2x+r \sin y \cos 2x=r\sin (2x+y)$$ so that $A=r\cos y, B=r\sin y$, we have $r^2=A^2+B^2$ and $\tan y=\frac BA$.

In the case in the question $r^2=2$ and $\tan y=1$ so we can have $y=\frac {\pi}4$. We reduce then to $$\sin \left(2x+\frac {\pi}4\right)=-\frac {\sqrt 2}2$$ which is then easy to solve.

1

Use sum to product for $\sin x + \sin y$ (it works also for $\cos x + \cos y$):

$$\begin{align} \sin 2x + \cos 2x&=\sin 2x + \sin\left(\frac{\pi}{2}-2x\right)\\ &=2\sin\frac{2x+\pi/2-2x}{2}\cos\frac{2x-\pi/2+2x}{2}\\ &=2\sin\frac{\pi}{4}\cos\left(2x-\frac{\pi}{4}\right)\\ &=\sqrt 2\cos\left(2x-\frac{\pi}{4}\right) \end{align}$$

The rest should be easy.

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I would use the old polar-to-Cartesian $x,y$ coordinates trick, but you are already using the variable $x,$ so I'll use a $t,y$ Cartesian coordinate plane instead with the $t$ coordinate horizontal (instead of $x$) and the $y$ coordinate vertical.

Let $t = \cos (2x)$ and let $y = \sin(2x).$ Then $t^2 + y^2 = 1,$ that is, $(t,y)$ is on the unit circle centered at the origin. We can also say that $(t,y)$ is the Cartesian coordinates of a point whose polar coordinates are $(r,\theta) = (1,2x).$

But you are given that $t + y = -1.$ Therefore $(t,y)$ is on the line $y = -1 - t.$ Find the points where that line intersects the unit circle; there are two such points. (A graph should make the points very easy to find in this particular case.)

Figure out the polar coordinates of those two points, and then remember that the polar coordinates $(r,\theta)$ are also equal to $(1,2x).$ Then solve for $x.$

0

As an alternative viewpoint to momo's answer. Write,

$$\langle \cos 2x, \sin 2x \rangle \cdot \langle 1,1 \rangle=-1$$

The first vector is a unit vector, the second has length $\sqrt{1^2+1^2}$. The angle between the vectors is $2x-\frac{\pi}{4}$ or $\frac{\pi}{4}-2x$ or etc. But it doesn't matter which one we choose because cosine is even. In fact you can check the resulting equation is exactly equal to the previous.

$$=\sqrt{1^2+1^2}\cos (2x-\frac{\pi}{4})=-1$$

-1

From $\sin 2x + \cos 2x =-1$ we get with $t=2x$:

$1=\sin^2 t+\cos^2 t+2 \sin t \cos t=1+ \sin(2t)$ .

Hence $ \sin(4x)=0$.

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    You can't say that $1 = {(\sin 2 x + \cos 2 x)}^2$ from $\sin 2 x + \cos 2 x = - 1$ because $t \mapsto t^2$ isn't injective on $\mathbb{R}$. – 2017-02-09
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    @joseabp91 If you rethink the equation as a function $f(x)=1+\sin(2x)+\cos(2x)$ then the zeros are the same as for $g(x)=f^2(x)$. Hence, I would think that Freds answer is correct. – 2017-02-09
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    @MrYouMath: the statement $$ 1 + \sin 2 x + \cos 2 x = 0 \Longleftrightarrow {(1 + \sin 2 x + \cos 2 x)}^2 = 0 $$ is correct, but the other statement $$ \sin 2 x + \cos 2 x = - 1 \Longleftrightarrow {(\sin 2 x + \cos 2 x)}^2 = 1 $$ is not correct: $$ \sin 2 x + \cos 2 x = - 1 \Rightarrow {(\sin 2 x + \cos 2 x)}^2 = 1 $$ is true, but $$ {(\sin 2 x + \cos 2 x)}^2 = 1 \Rightarrow \sin 2 x + \cos 2 x = - 1 $$ is clearly false (take for example $x = \pi$). Remember we want to find $x \in \mathbb{R}$ such that $\sin 2 x + \cos 2 x = - 1$, not such that ${(\sin 2 x + \cos 2 x)}^2 = 1$ – 2017-02-09
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    @José From $a=-1$ we get $a^2=1$. The downvote is not o.k. ! – 2017-02-09
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    @Fred: if $a = - 1$, then $a^2 = 1$, but you can't say that $a = - 1$ if $a^2 = 1$ because $a$ can be nonnegative, right? – 2017-02-09
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    The implication only goes one way, in $a^2 = 1$ you get *both solutions* $a = -1$ and $a = 1$, while only one is valid for the original formula. – 2017-02-09
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    I think the Fred's answer says that $$ \sin 4 x = 0 \Longrightarrow \sin 2 x + \cos 2 x = - 1 $$ but it is not correct (take $x = 0$). – 2017-02-09
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    No. My answer is the reversed Implikation. – 2017-02-09
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    Final equation indeed has more roots than initial, but all roots of initial are roots of the final version. So correct approach is to solve final equation and then check all roots using initial equation. Resulted solution is all roots left after removing irrelevant ones. So Fred's approach is good considering filtering results. I suggest Fred to improve answer with corresponding note, then downvote should be removed. – 2017-02-09
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    @Fred: if your answer is the reversed implication, then it is not useful for this question. I repeat the argument again: we want to find $x \in \mathbb{R}$ such that $\sin 2 x + \cos 2 x = - 1$, not such that $\sin 4 x = 0$. – 2017-02-09
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    There is an equation in the question, so we have to give all the solutions. In fact, if $\sin 2 x + \cos 2 x = - 1$, then $\sin 4 x = 0$, but it doesn't answer to the question. – 2017-02-09
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    @joseabp91, it does answer to the question once you filtered out irrelevant roots using initial equation. It should be mentioned in answer, but it's a fine way to get correct solution. – 2017-02-09
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    If you filter many solutions yes but in my opinion it's more complicate than look for a new methode to solve the initial equation. – 2017-02-09
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    In my answer I showed that $\sin 2x + \cos 2x =-1$ implies $\sin(4x)=0$. I am aware that not every solution of $\sin(4x)=0$ is a solution of $\sin 2x + \cos 2x =-1$. My intention was, that an interested reader should invest a little bit (a very, very litle bit) effort to investigate, which solutions of $\sin(4x)=0$ are also solutions of $\sin 2x + \cos 2x =-1$. I appologize that I encumbered some readers efforts which are to hard ! – 2017-02-10
-1

It's possible to represent sum of sine and cosine of the same argument as follows

$$\sin t + \cos t = k \cdot \sin (t + \pi/4)$$

($t = 2x$ for this task), where $k$ can be found from equation above by using "sine of sum of angles" identity on the right part. So in your case initial equation will be brought to much easier one:

$$k \cdot \sin (2x + \pi/4) = -1$$