Trig Equations

[Mycroft]

Hello!
I'm having a really hard time with solving trig equations. Here is one I have been working on and keep getting wrong: cos(2x-pi/2)=-1
I'm told to solve it on the interval [0,2pi].
So far, I have found that 2x-pi/2=pi, because the cos(something)=-1 gives pi. Then I solved for x and got that x=3pi/4. My book also lists 7pi/4 as the answer. I thought it would have been 5pi/4 because if it was originally a problem where cosine was negative. What am I doing wrong?
Thanks!
~Mycroft

 

[Mycroft]

Oh and I hope this was posted in the right section... trig, not precalc, right? I don't know if it matters too much, but sorry if it does!

 

[soroban]

Hello, Mycroft!

The "2" in \(\displaystyle 2x\) can cause problems . . .

\(\displaystyle \cos\left(2x-\frac{\pi}{2}\right)\,=\,-1\;\;\text{ for }x \in [0,\,2\pi]\)

You are correct: .\(\displaystyle \cos(\pi) \,=\,-1\)
. . But you should write the general form of the answer.

\(\displaystyle \cos\left(2x - \frac{\pi}{2}\right) \:=\:-1 \quad\Rightarrow\quad 2x - \frac{\pi}{2} \:=\: \pi + 2\pi n\)

. . \(\displaystyle 2x \;=\;\frac{3\pi}{2} + 2\pi n \quad\Rightarrow\quad x \;=\;\frac{3\pi}{4} + \pi n\)


And for \(\displaystyle n = 0\text{ and }1,\)
. . we have: .\(\displaystyle x \:=\:\dfrac{3\pi}{4}\text{ and }\dfrac{7\pi}{4}\) . . . both are in the interval.

 

[Deleted member 4993]

Hello!
I'm having a really hard time with solving trig equations. Here is one I have been working on and keep getting wrong: cos(2x-pi/2)=-1
I'm told to solve it on the interval [0,2pi].
So far, I have found that 2x-pi/2=pi, because the cos(something)=-1 gives pi. Then I solved for x and got that x=3pi/4. My book also lists 7pi/4 as the answer. I thought it would have been 5pi/4 because if it was originally a problem where cosine was negative. What am I doing wrong?
Thanks!
~Mycroft

If you put x = 5(pi)/4 into the expression you get:

cos(2*5*(pi)/4 - pi/2) = cos(4(pi)/2) = +1

does not satisfy original condition.

 

[Mycroft]

Thanks guys! That makes sense, I forgot that you generalize the answer and then get x by itself, so you end up dividing the two out of 2pin. This is complicated. Thanks again!