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I'm trying to find the points of intersection of $\sin x$ and $\cos x$ between $0$ and $2\pi$. I've tried but I keep getting 4 solutions... Would someone please be able to take me through the process?

I squared $\sin x$ and $\cos x$ and then got $\sin x = \frac{\pm 1}{\sqrt{2}}$ which gave the solutions $\pi\over 4$, $3\pi \over 4$, $5\pi \over 4$ and $7\pi\over 4$.

MonkeyKing
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Bob
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    Would you please show your work? (Take a picture if you have to if you're having trouble writing it up.) Sometimes when you're doing algebraic problems you end up getting "spurious" solutions that solve some equation you obtained later on down the line but are not actually solutions to the original equation. This happens when you square an equation, for instance. – Cameron Williams May 08 '15 at 04:00
  • Which four do you get? –  May 08 '15 at 04:01
  • Thanks for editing to add your solutions! Why don't you try plugging each of the four of those into your original equation to see if all of them solve it? (A simpler way to do this problem is to use the fact that $\sin x = \cos x$ gives that $\tan x = 1$.) – Cameron Williams May 08 '15 at 04:05
  • hint: given and angle $x$ there corresponds a point on the unit circle whose coordinates are $(\cos x, \sin x)$. Where are the horizontal and vertical components equal to each other? – John Joy May 08 '15 at 14:49

2 Answers2

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$\sin x = \cos x \implies \sin^2x = \cos^2 x $ together with the proviso that $\sin x$ and $\cos x$ must share the same sign.

So the only two solutions are

$\sin x=\frac{1}{\sqrt 2}$, $\cos x=\frac{1}{\sqrt 2}$ giving $x=\frac{\pi}{4}$

and

$\sin x=\frac{-1}{\sqrt 2}$, $\cos x=\frac{-1}{\sqrt 2}$ giving $x=\frac{5\pi}{4}$

WW1
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There are 2 solutions. $$\sin x = \cos x \Rightarrow \tan x = 1$$ because clearly $\cos x \neq 0$ so we can divide both sides by $\cos x$.

Therefore, when $x \in [0,2\pi]$, the solutions are $x_1 = \frac{1}{4}\pi$ and $x_2 = \frac{5}{4}\pi$.

Here is the figure.

Edit: When you square both sides, you will make $(-\sin x)^2 = (\sin x)^2$. If $-\sin x = \cos x$, their squares are equal, which is a problem since they are not $0$.

MonkeyKing
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