Solving trigonometric equations

**TI-84** · February 26th 2008, 05:02 PM

Solve the trigonometric equation analytically for values of $x$ for $0\le x <2\pi$

$sin\left(x - \frac{\pi}{4}\right) = cos\left(x - \frac{\pi}{4}\right)$

$sin\;x\;cos\;\frac{\pi}{4} - cos\;x\;sin\;\frac{\pi}{4} = cos\;x\;cos\;\frac{\pi}{4} + sin\;x\;sin\;\frac{\pi}{4}$

$sin\;x\;\left(\frac{\sqrt{2}}{2}\right) - cos\;x\;\left(\frac{\sqrt{2}}{2}\right) = cos\;x\;\left(\frac{\sqrt{2}}{2}\right) + sin\;x\;\left(\frac{\sqrt{2}}{2}\right)$

$2\;cos\;x\;\left(\frac{\sqrt{2}}{2}\right)$

If what I did so far is right. I'm stuck at this point.

**mr fantastic** · February 26th 2008, 05:19 PM

Originally Posted by TI-84

Solve the trigonometric equation analytically for values of $x$ for $0\le x <2\pi$

$sin\left(x - \frac{\pi}{4}\right) = cos\left(x - \frac{\pi}{4}\right)$

$sin\;x\;cos\;\frac{\pi}{4} - cos\;x\;sin\;\frac{\pi}{4} = cos\;x\;cos\;\frac{\pi}{4} + sin\;x\;sin\;\frac{\pi}{4}$

$sin\;x\;\left(\frac{\sqrt{2}}{2}\right) - cos\;x\;\left(\frac{\sqrt{2}}{2}\right) = cos\;x\;\left(\frac{\sqrt{2}}{2}\right) + sin\;x\;\left(\frac{\sqrt{2}}{2}\right)$

$2\;cos\;x\;\left(\frac{\sqrt{2}}{2}\right)$

If what I did so far is right. I'm stuck at this point.

You're stuck because your last line should read

$2\;cos\;x\;\left(\frac{\sqrt{2}}{2}\right)$ = 0,

that is

$\sqrt{2}\;\cos x$ = 0,

**TI-84** · February 26th 2008, 05:30 PM

But when I bring everything over to the right side to isolate x doesnt it just become x=0? the books answer is $\frac{\pi}{2}\;and\;\frac{3\pi}{2}$

**mr fantastic** · February 26th 2008, 05:40 PM

Originally Posted by mr fantastic

You're stuck because your last line should read

$2\;cos\;x\;\left(\frac{\sqrt{2}}{2}\right)$ = 0,

that is

$\sqrt{2}\;\cos x$ = 0,

Next step: Divide both sides by $\sqrt{2}$ :

$\cos x = 0$

Therefore x = ......

**mr fantastic** · February 26th 2008, 05:43 PM

Originally Posted by mr fantastic

Next step: Divide both sides by $\sqrt{2}$ :

$\cos x = 0$

Therefore x = ......

Perhaps I should write it as $\cos (x) = 0$ .

Hmmmmmm ....... You do understand that $\cos (x) = 0$ means cos of x, NOT $cos \times x$ (which has no meaning.

In the same way that $\sqrt{2}$ means the square root of 2, not square root times 2 ......

So you're looking for values of x such that the cos of those values is equal to 0 ......

**TI-84** · February 26th 2008, 08:34 PM

$<br /> 2\;cos\;x\;\left(\frac{\sqrt{2}}{2}\right) = 0<br />$

$cos\;x\;\left(\frac{\sqrt{2}}{2}\right) = 0$

$cos\;x = 0$

$x = Cos^{-1}\;(0)$

$x=90$ and $x=270$

Is that correct?

**mr fantastic** · February 27th 2008, 03:28 AM

Originally Posted by TI-84

$<br /> 2\;cos\;x\;\left(\frac{\sqrt{2}}{2}\right) = 0<br />$

$cos\;x\;\left(\frac{\sqrt{2}}{2}\right) = 0$

$cos\;x = 0$

$x = Cos^{-1}\;(0)$

$x=90$ and $x=270$

Is that correct?

Yes. You can substitute into the original equation and confirm this.

Edit: Seeing Soroban's reply reminds me - your domain is in radians, so your solutions for x need to be in radians. So convert your answer from degrees into radians (and get what Soroban has got using a much easier method).

**Soroban** · February 27th 2008, 04:34 AM

Hello, TI-84

Solve for $x,\;\;0 \le x <2\pi$

. . $\sin\left(x - \frac{\pi}{4}\right) \:= \:\cos\left(x - \frac{\pi}{4}\right)$

Divide both sides by $\cos\left(x-\frac{\pi}{4}\right)$

. . $\frac{\sin\left(x-\frac{\pi}{4}\right)}{\cos\left(x-\frac{\pi}{4}\right)} \;=\;1 \quad\Rightarrow\quad \tan\left(x-\frac{\pi}{4}\right) \;=\;1$

Hence: . $x - \frac{\pi}{4} \;=\;\left\{\frac{\pi}{4},\:\frac{5\pi}{4}\right\} \quad\Rightarrow\quad x \;=\;\left\{\frac{\pi}{2},\:\frac{3\pi}{2}\right\}$

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