# Thread: Help solving trig equation 1

1. ## Help solving trig equation 1

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

2. ## recall that

Originally Posted by dorkymichelle
my ugly attempt
$\displaystyle cos(2x-\pi/2) =\sqrt2/2$
$\displaystyle cos2xcos\pi/2+sin2xsin\pi/2 = \sqrt2/2$
$\displaystyle (2cos^2-1)(cos\pi/2)+(2sinxcosx)(sin\pi/2)=\sqrt2/2$
/edit got another step
$\displaystyle (2cos^2-1)(1)0+(2sinxcosx)(1)=\sqrt2/2$
$\displaystyle (2sinxcosx)=\sqrt2/2$
recall that $\displaystyle cos\left(\frac{\pi}{2}\right) = 0$
and $\displaystyle sin\left(\frac{\pi}{2}\right) = 1$

so

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

can you finish from here

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

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

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

In the answers, it is $\displaystyle 3\pi/8+n\pi,9\pi/8+n\pi$ (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 $\displaystyle sin\left(2x\right) = \frac{\sqrt{2}}{2}$

then

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

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

so

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

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

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

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

5. Thankyou!

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

then

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

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

so

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

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

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

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 $\displaystyle \frac{3\pi}{4}$

7. 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?

8. 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?
$\displaystyle \frac{\pi}{8} +\pi n$

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

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

find common denominators

$\displaystyle \frac{\pi}{8} +\frac{8\pi}{8}= \frac{9\pi}{8}$

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

10. 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.

The correct way to write your answer is

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

You have to becareful some times because they may put a constraint on what your solutions may be.

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

12. Originally Posted by dorkymichelle
The question told me to find all solutions
The teacher's answer on the study guide is
$\displaystyle 3\pi/8+n\pi and 9\pi/8 +n\pi$
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

13. Thankyou so much. i fainlly understand this. The differnt answers were confusing me day.

14. Originally Posted by dorkymichelle
Thankyou so much. i fainlly understand this. The differnt answers were confusing me day.
Glad I could help