# Help solving trig equation 1

• Dec 8th 2009, 09:23 PM
dorkymichelle
Help solving trig equation 1
my ugly attempt
$cos(2x-\pi/2) =\sqrt2/2$
$cos2xcos\pi/2+sin2xsin\pi/2 = \sqrt2/2$
$(2cos^2-1)(cos\pi/2)+(2sinxcosx)(sin\pi/2)=\sqrt2/2$
/edit got another step
$(2cos^2-1)(1)0+(2sinxcosx)(1)=\sqrt2/2$
$(2sinxcosx)=\sqrt2/2$
• Dec 8th 2009, 09:43 PM
bigwave
recall that
Quote:

Originally Posted by dorkymichelle
my ugly attempt
$cos(2x-\pi/2) =\sqrt2/2$
$cos2xcos\pi/2+sin2xsin\pi/2 = \sqrt2/2$
$(2cos^2-1)(cos\pi/2)+(2sinxcosx)(sin\pi/2)=\sqrt2/2$
/edit got another step
$(2cos^2-1)(1)0+(2sinxcosx)(1)=\sqrt2/2$
$(2sinxcosx)=\sqrt2/2$

recall that $cos\left(\frac{\pi}{2}\right) = 0$
and $sin\left(\frac{\pi}{2}\right) = 1$

so

$cos2x(0)+sin2x(1) = \sqrt2/2$

can you finish from here
• Dec 8th 2009, 09:52 PM
dorkymichelle
I ended up with
$(2sinxcosx)=\sqrt2/2$
what is the next methodlical step?(i can't spell..)

In the answers, it is $3\pi/8+n\pi,9\pi/8+n\pi$ (its a study guide for my next test)
How can I solve an equation without a basic unit circle trig. value?
• Dec 8th 2009, 11:11 PM
bigwave
Quote:

Originally Posted by dorkymichelle
I ended up with
$(2sinxcosx)=\sqrt2/2$
what is the next methodlical step?(i can't spell..)

In the answers, it is $3\pi/8+n\pi,9\pi/8+n\pi$ (its a study guide for my next test)
How can I solve an equation without a basic unit circle trig. value?

lets go back to $sin\left(2x\right) = \frac{\sqrt{2}}{2}$

then

$
sin^{-1}\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4}
$

therefore $2x = \frac{\pi}{4}$

so

$x = \frac{\pi}{8}$

if you plug this back in the original it will ck out

if you add $\pi$ to $\frac{\pi}{8}$ you get $\frac{9\pi}{8}$
$n\pi$where $n$ is an interger
this is because $sin(x)$ has a period of $\pi$

it is a unit circle value just not a commonly known one.
• Dec 9th 2009, 08:24 PM
dorkymichelle
Thankyou!
• Dec 9th 2009, 08:35 PM
11rdc11
Quote:

Originally Posted by bigwave
lets go back to $sin\left(2x\right) = \frac{\sqrt{2}}{2}$

then

$
sin^{-1}\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4}
$

therefore $2x = \frac{\pi}{4}$

so

$x = \frac{\pi}{8}$

if you plug this back in the original it will ck out

if you add $\pi$ to $\frac{\pi}{8}$ you get $\frac{9\pi}{8}$
$n\pi$where $n$ is an interger
this is because $sin(x)$ has a period of $\pi$

it is a unit circle value just not a commonly known one.

Correct but you must also remember that sin is positive in the second quad at $\frac{3\pi}{4}$
• Dec 9th 2009, 09:19 PM
dorkymichelle
hmm.. if I also use 3\pi/4 then x would = to 3\pi/8 +n\pi, that is on the answer, but \pi/8 +n\pi is not on the answer.
And also, I don't understand how you get 9\pi/8?
Did he just skip the 2st quadrant, and wrote the 2nd answer as going half a circle from the 2nd quadrant instead of using the answer from the 1st?
• Dec 9th 2009, 09:24 PM
11rdc11
Quote:

Originally Posted by dorkymichelle
hmm.. if I also use 3\pi/4 then x would = to 3\pi/8 +n\pi, that is on the answer, but \pi/8 +n\pi is not on the answer.
And also, I don't understand how you get 9\pi/8?
Did he just skip the 2st quadrant, and wrote the 2nd answer as going half a circle from the 2nd quadrant instead of using the answer from the 1st?

$\frac{\pi}{8} +\pi n$

n just stands for an interger, n=1,2,3,4,...

$\frac{\pi}{8} +(\pi)(1)$

find common denominators

$\frac{\pi}{8} +\frac{8\pi}{8}= \frac{9\pi}{8}$
• Dec 9th 2009, 09:31 PM
dorkymichelle
So basicly, if I put \pi/8,3\pi/8 and 9\pi/8 they will all be correct answers?
• Dec 9th 2009, 09:37 PM
11rdc11
Quote:

Originally Posted by dorkymichelle
So basicly, if I put \pi/8,3\pi/8 and 9\pi/8 they will all be correct answers?

Yes since trig equations are periodic.

$\frac{\pi}{8} +\pi n, \frac{3\pi}{8} + \pi n$

You have to becareful some times because they may put a constraint on what your solutions may be.
• Dec 9th 2009, 10:00 PM
dorkymichelle
The question told me to find all solutions
The teacher's answer on the study guide is
$3\pi/8+n\pi and 9\pi/8 +n\pi$
so I guess both ways is correct, but my teacher is just picky about solutions?
• Dec 9th 2009, 10:52 PM
11rdc11
Quote:

Originally Posted by dorkymichelle
The question told me to find all solutions
The teacher's answer on the study guide is
$3\pi/8+n\pi and 9\pi/8 +n\pi$
so I guess both ways is correct, but my teacher is just picky about solutions?

Correct they are both correct and will give you all the solutions
• Dec 9th 2009, 11:08 PM
dorkymichelle
Thankyou so much. i fainlly understand this. The differnt answers were confusing me day.
• Dec 9th 2009, 11:15 PM
11rdc11
Quote:

Originally Posted by dorkymichelle
Thankyou so much. i fainlly understand this. The differnt answers were confusing me day.