solve the following equations on (0 less than or equal to theta less than or equal to 2 pi)
1.)tan 2 theta=1
2.)sin 2theta=square root 2costheta
for #1 i get pi/3 and 5pi/3
and for #2 pi/4 but am not sure if theye are correct

Question

zzr0ck3r · Answer

1)
let x = cos(theta)
2x^2+x=0
x(2x+1) = 0
so
cos(theta) = 0 and 2cos(theta) + 1 = 0 ---> cos(theta) = -1/2
so
solve these
cos(theta) = 0 and cos(theta) = -1/2 between 0 and 2pi
this is true when theta = pi/2, 3pi/2, 4pi/6, and 8pi/6

campbell_st · Answer

so is the 1st question 

1. tan(2x) = 1 

or

$$2. ... \tan^2(x) = 1$$

anonymous · Answer

the first one

anonymous · Answer

i made a mistake when i wrote the answers i got for that problem with another problem that   had a tan theta in it but the pi/8 is what i got

campbell_st · Answer

ok... so its easy 

so tan is positive in the 1st and 3rd quadrants

and you know 

$$2x = \tan^{-1}(1) .......or..... 2x = \frac{\pi}{4}$$

so find x

1st quadrant the the answer you find. 

3rd quadrant is $$\pi + answer$$

campbell_st · Answer

hope that helps

campbell_st · Answer

well you are correct with 1 answer, but there are 2 in the given domain.,

campbell_st · Answer

and for the 2nd equation is it 

$$\sin(2x) = \sqrt{2}\cos(x)$$

anonymous · Answer

yeah thats the second equation

anonymous · Answer

so u say 2x=pi/4 wud i divide both sides by 2 to get just x giving me 1/8?

anonymous · Answer

and then i just add pi to 1/8?

campbell_st · Answer

well actually 

$$x = \frac{\pi}{8}$$

the other answer is 

$$x = \pi + \frac{\pi}{8}....or......x = \frac{\pi}{8}$$

campbell_st · Answer

opps should be 

$$x = \frac{9\pi}{8}$$

campbell_st · Answer

so 2 answers to the question

anonymous · Answer

ohh my bad yeah i ment pi/8 but i just didnt put the pi sorry

campbell_st · Answer

so looking at the 2nd problem 

sin(2x) = 2sin(x)cos(x) 

so you need this substitution for the 2nd equation..

campbell_st · Answer

so its

$$2 \sin(x)\cos(x) = \sqrt{2}\cos(x)$$

divide both sides by 2cos(x) gives

$$\sin(x) = \frac{\sqrt{2}}{2}$$

this is an exact value. 

so 

$$x = \frac{\pi}{4}$$

again because of the domain. there are 2 solutions

1st quadrant, as shown above

and 2nd quadrant which uses 

$$\pi - answer$$

hope this helps

anonymous · Answer

so 3pi/4 is the second answer?

campbell_st · Answer

I'd say so

zzr0ck3r · Answer

so confused...did the question get edited and changed? Or did I just make up a problem and solve it?

anonymous · Answer

no lol i edited the question n removed the one u had helped me solve

anonymous · Answer

thank for your help tho =D ... i have final thursday on doing problems like this n im so confused on how to do it n i gotta pass or else i fail the class since i didnt do so well on the test =[

zzr0ck3r · Answer

gl