MTALHAHASSAN2
  • MTALHAHASSAN2
Need help!! 8. Solve for theta. 0
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
campbell_st
  • campbell_st
factor out cos \[\cos(\theta)(5\sin(\theta) -3) \] now set each factor to zero and solve \[\cos(tehta) = 0~~~and~~~5\sin(\theta) - 3 = 0\]
MTALHAHASSAN2
  • MTALHAHASSAN2
|dw:1434604876572:dw|
MTALHAHASSAN2
  • MTALHAHASSAN2
can someone plz help me

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

MTALHAHASSAN2
  • MTALHAHASSAN2
@Michele_Laino
MTALHAHASSAN2
  • MTALHAHASSAN2
@KyanTheDoodle
MTALHAHASSAN2
  • MTALHAHASSAN2
@RezForLife
MTALHAHASSAN2
  • MTALHAHASSAN2
@Alexander95
MTALHAHASSAN2
  • MTALHAHASSAN2
@bleebleebleed
MTALHAHASSAN2
  • MTALHAHASSAN2
@Compassionate
MTALHAHASSAN2
  • MTALHAHASSAN2
@Data_LG2
MTALHAHASSAN2
  • MTALHAHASSAN2
@esam2
MTALHAHASSAN2
  • MTALHAHASSAN2
@freckles
anonymous
  • anonymous
From your unit circle you can pick out where cosine is 0. For the sine solutions, you already have sin Θ = 0.6. Use the sin^-1 button on your calculator to find Θ Θ = sin^-1 (0.6) This gives an angle in the 1st quadrant. sine is also positive in the 2nd quadrant. Subtract the other solution from pi to get the angle in the 2nd quadrant. Θ = π - sin^-1 (0.6)
anonymous
  • anonymous
do still need help i would have messaged u but i got blocked for god knows what
anonymous
  • anonymous
@MTALHAHASSAN2
MTALHAHASSAN2
  • MTALHAHASSAN2
Θ = sin^-1 (0.6) i get 37 degree
anonymous
  • anonymous
ican help or fleather me
MTALHAHASSAN2
  • MTALHAHASSAN2
@SkaterBoyShawn what is ur problem
MTALHAHASSAN2
  • MTALHAHASSAN2
I don't need ur help
MTALHAHASSAN2
  • MTALHAHASSAN2
@Michele_Laino
anonymous
  • anonymous
i have no prob u said fleather me then blocked me i am confuesd make sense i dont know what i did to u
UsukiDoll
  • UsukiDoll
inb4 crazy ban from original poster ...stop mass tagging people.
MTALHAHASSAN2
  • MTALHAHASSAN2
@SkaterBoyShawn stop
anonymous
  • anonymous
yes 37° is one solution. 180° - 37° = 143° is another solution
anonymous
  • anonymous
now what did i do
MTALHAHASSAN2
  • MTALHAHASSAN2
wait but at the back the answer is different
MTALHAHASSAN2
  • MTALHAHASSAN2
it is something like 74
anonymous
  • anonymous
stop tagging me iam bothing u so leave me alone bye have a good evning
Michele_Laino
  • Michele_Laino
we can factor out cos( \theta), so we can write: \[\Large \cos \theta \left( {5\sin \theta - 3} \right) = 0\]
campbell_st
  • campbell_st
you have missed on of the key aspects of the question, you are working in radians and you should have 4 angles as the solution
MTALHAHASSAN2
  • MTALHAHASSAN2
no@campbell
MTALHAHASSAN2
  • MTALHAHASSAN2
the answer is given in degrees
anonymous
  • anonymous
@MTALHAHASSAN2 yes he's right. It is in radians
campbell_st
  • campbell_st
you solutions to the factor are correct.. \[\cos(\theta) = 0\] what 2 angles measure make this true, remember its in radians...
anonymous
  • anonymous
"0
MTALHAHASSAN2
  • MTALHAHASSAN2
oh yeah
MTALHAHASSAN2
  • MTALHAHASSAN2
my bad
campbell_st
  • campbell_st
the domain is given as \[0 < \theta l< 2\pi\] so you should answer in radians
MTALHAHASSAN2
  • MTALHAHASSAN2
yep the answer is in radian
campbell_st
  • campbell_st
so the solutions for \[\cos(\theta) = 0\] should be straight forward the 2 angles for \[\sin(\theta) = \frac{3}{5}\] are in the 1st quadrant and 2nd quadrant
imqwerty
  • imqwerty
5sin(θ)cos(θ)-3cos(θ) = 0 divide by 5 sin(θ)cos(θ) -3/5cos(θ)=0 3/5=sin37(approx) sin(θ)cos(θ)-sin(37)cos(θ)=0 cos(θ)[sin(θ)-sin(37)]=0 either cos(θ)=0 i.e, (θ)=90 or sin(θ)-sin(37)=0 i.e (θ)=37 or 143 :)
campbell_st
  • campbell_st
does that make sense..?
MTALHAHASSAN2
  • MTALHAHASSAN2
so 37 and 143 be the answer
MTALHAHASSAN2
  • MTALHAHASSAN2
but i have to convert that in radians
campbell_st
  • campbell_st
so find the 1st quadrant angle for theta and then 2nd quadrant is \[\pi - \theta\]
campbell_st
  • campbell_st
this online calculator will do it for you http://web2.0calc.com/ just select radians
campbell_st
  • campbell_st
then input the ratio
MTALHAHASSAN2
  • MTALHAHASSAN2
how
campbell_st
  • campbell_st
select radian press 2nd... then asin which is arcsin \[asin = \sin^{-1}\] then enter 0.6 and = what do you get?
campbell_st
  • campbell_st
oops should select Rad.. bottom left corner
MTALHAHASSAN2
  • MTALHAHASSAN2
ok
campbell_st
  • campbell_st
then 2nd asin (0.6) =
MTALHAHASSAN2
  • MTALHAHASSAN2
0.605
imqwerty
  • imqwerty
the solutions are 90 = pi/2 rad 37=37pi/180 rad 143=143pi/180 rad do the calculations and get the answer in radians
campbell_st
  • campbell_st
ummm not quite... I got 0.6435011.... can you just check again you have to press 2nd to before closing the brackets then =
MTALHAHASSAN2
  • MTALHAHASSAN2
i goted 0.64
campbell_st
  • campbell_st
great so that is the 1st quadrant answer to the nearest hundredth now the 2nd quadrant is \[\pi - 0.64\]
MTALHAHASSAN2
  • MTALHAHASSAN2
2.50
MTALHAHASSAN2
  • MTALHAHASSAN2
right
MTALHAHASSAN2
  • MTALHAHASSAN2
@campbell_st
campbell_st
  • campbell_st
yes.. that's it... well done
MTALHAHASSAN2
  • MTALHAHASSAN2
wait but at the back they have four solution
MTALHAHASSAN2
  • MTALHAHASSAN2
these two are the same but we are missing two
campbell_st
  • campbell_st
ok... so they are the values for sin now the values of cos you have \[\cos(\theta) = 0\] what angles make it true..?
MTALHAHASSAN2
  • MTALHAHASSAN2
90
campbell_st
  • campbell_st
here is the curve |dw:1434608602050:dw|
campbell_st
  • campbell_st
remember its in radians
MTALHAHASSAN2
  • MTALHAHASSAN2
pie/2
campbell_st
  • campbell_st
so where does the cos curve cut the horizontal axis..?
campbell_st
  • campbell_st
great, that's 1 and the other is..?
MTALHAHASSAN2
  • MTALHAHASSAN2
270 which is 3pie/2
campbell_st
  • campbell_st
that's correct... so you have 4 angles...
MTALHAHASSAN2
  • MTALHAHASSAN2
yes thanks a lot
campbell_st
  • campbell_st
glad to help

Looking for something else?

Not the answer you are looking for? Search for more explanations.