solve $3sintheta+4costheta=0$
up vote
1
down vote
favorite
Solve for $0 < theta < 360$
Question
$3 sin theta + 4 cos theta = 0$
Please help. I really can't figure this out
Thanks :)
What I have tried
I tried using the a $sin theta + b cos theta = r sin (theta + a)$ rule but didn't work out (or I might have done it wrong)
trigonometry
edited Jul 14 '14 at 1:53
Edward
305412
asked Jul 14 '14 at 1:44
user164215
62
add a comment |
up vote
1
down vote
favorite
Solve for $0 < theta < 360$
Question
$3 sin theta + 4 cos theta = 0$
Please help. I really can't figure this out
Thanks :)
What I have tried
I tried using the a $sin theta + b cos theta = r sin (theta + a)$ rule but didn't work out (or I might have done it wrong)
trigonometry
edited Jul 14 '14 at 1:53
Edward
305412
asked Jul 14 '14 at 1:44
user164215
62
Hint: convert to a ratio relationship.
– Gina
Jul 14 '14 at 1:46
@Gina Why don't you make that an answer? (Maybe be a little more explicit.)
– Cameron Williams
Jul 14 '14 at 1:50
Which do you use 360 (rad), or 360 (degree)?
– choco_addicted
Nov 29 at 6:26
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Solve for $0 < theta < 360$
Question
$3 sin theta + 4 cos theta = 0$
Please help. I really can't figure this out
Thanks :)
What I have tried
I tried using the a $sin theta + b cos theta = r sin (theta + a)$ rule but didn't work out (or I might have done it wrong)
trigonometry
edited Jul 14 '14 at 1:53
Edward
305412
asked Jul 14 '14 at 1:44
user164215
62
Solve for $0 < theta < 360$
Question
$3 sin theta + 4 cos theta = 0$
Please help. I really can't figure this out
Thanks :)
What I have tried
I tried using the a $sin theta + b cos theta = r sin (theta + a)$ rule but didn't work out (or I might have done it wrong)
trigonometry
trigonometry
edited Jul 14 '14 at 1:53
Edward
305412
asked Jul 14 '14 at 1:44
user164215
62
edited Jul 14 '14 at 1:53
Edward
305412
asked Jul 14 '14 at 1:44
user164215
62
edited Jul 14 '14 at 1:53
Edward
305412
edited Jul 14 '14 at 1:53
Edward
305412
edited Jul 14 '14 at 1:53
Edward
305412
305412
asked Jul 14 '14 at 1:44
user164215
62
asked Jul 14 '14 at 1:44
user164215
62
asked Jul 14 '14 at 1:44
user164215
62
62
Hint: convert to a ratio relationship.
– Gina
Jul 14 '14 at 1:46
@Gina Why don't you make that an answer? (Maybe be a little more explicit.)
– Cameron Williams
Jul 14 '14 at 1:50
Which do you use 360 (rad), or 360 (degree)?
– choco_addicted
Nov 29 at 6:26
add a comment |
Hint: convert to a ratio relationship.
– Gina
Jul 14 '14 at 1:46
@Gina Why don't you make that an answer? (Maybe be a little more explicit.)
– Cameron Williams
Jul 14 '14 at 1:50
Which do you use 360 (rad), or 360 (degree)?
– choco_addicted
Nov 29 at 6:26
Hint: convert to a ratio relationship.
– Gina
Jul 14 '14 at 1:46
Hint: convert to a ratio relationship.
– Gina
Jul 14 '14 at 1:46
@Gina Why don't you make that an answer? (Maybe be a little more explicit.)
– Cameron Williams
Jul 14 '14 at 1:50
@Gina Why don't you make that an answer? (Maybe be a little more explicit.)
– Cameron Williams
Jul 14 '14 at 1:50
Which do you use 360 (rad), or 360 (degree)?
– choco_addicted
Nov 29 at 6:26
Which do you use 360 (rad), or 360 (degree)?
– choco_addicted
Nov 29 at 6:26
add a comment |
3 Answers
3
active
oldest
votes
up vote
2
down vote
$3sin theta +4costheta =0$
Notice that $theta=frac{pi}{2}+pi k$ is not a solution to this equation, and so it is valid to divide by $cos theta$:
$3tan theta =-4$, $tan theta =-frac{4}{3}$
And so $theta$ is....?
answered Jul 14 '14 at 1:55
Oria Gruber
6,38532360
add a comment |
up vote
0
down vote
HINT: divide the ecuation by $costheta$
answered Jul 14 '14 at 1:52
AsdrubalBeltran
2,8061916
add a comment |
up vote
0
down vote
Divide both sides by 5, and observe that the coefficients satisfy
$left(frac{3}{5}right)^2+ left(frac{4}{5}right)^2=1.$ Consequently there is some angle $alpha$ such that $cos alpha = 3/5, sinalpha =4/5$. (For comparison with other answers, note that $tanalpha = 4/3$). With this in mind we can use the sum-to-product identity to write
$$ sin(alpha+theta) = cosalpha sin theta+sinalpha costheta = frac{3}{5}sintheta+frac{4}{5}costheta = 0.$$ So all that's left is to find the solutions of this equation.
answered Jul 14 '14 at 2:08
Semiclassical
11k32464
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
$3sin theta +4costheta =0$
Notice that $theta=frac{pi}{2}+pi k$ is not a solution to this equation, and so it is valid to divide by $cos theta$:
$3tan theta =-4$, $tan theta =-frac{4}{3}$
And so $theta$ is....?
answered Jul 14 '14 at 1:55
Oria Gruber
6,38532360
add a comment |
up vote
2
down vote
$3sin theta +4costheta =0$
Notice that $theta=frac{pi}{2}+pi k$ is not a solution to this equation, and so it is valid to divide by $cos theta$:
$3tan theta =-4$, $tan theta =-frac{4}{3}$
And so $theta$ is....?
answered Jul 14 '14 at 1:55
Oria Gruber
6,38532360
add a comment |
up vote
2
down vote
up vote
2
down vote
$3sin theta +4costheta =0$
Notice that $theta=frac{pi}{2}+pi k$ is not a solution to this equation, and so it is valid to divide by $cos theta$:
$3tan theta =-4$, $tan theta =-frac{4}{3}$
And so $theta$ is....?
answered Jul 14 '14 at 1:55
Oria Gruber
6,38532360
$3sin theta +4costheta =0$
Notice that $theta=frac{pi}{2}+pi k$ is not a solution to this equation, and so it is valid to divide by $cos theta$:
$3tan theta =-4$, $tan theta =-frac{4}{3}$
And so $theta$ is....?
answered Jul 14 '14 at 1:55
Oria Gruber
6,38532360
answered Jul 14 '14 at 1:55
Oria Gruber
6,38532360
answered Jul 14 '14 at 1:55
Oria Gruber
6,38532360
answered Jul 14 '14 at 1:55
Oria Gruber
6,38532360
6,38532360
add a comment |
add a comment |
up vote
0
down vote
HINT: divide the ecuation by $costheta$
answered Jul 14 '14 at 1:52
AsdrubalBeltran
2,8061916
add a comment |
up vote
0
down vote
HINT: divide the ecuation by $costheta$
answered Jul 14 '14 at 1:52
AsdrubalBeltran
2,8061916
add a comment |
up vote
0
down vote
up vote
0
down vote
HINT: divide the ecuation by $costheta$
answered Jul 14 '14 at 1:52
AsdrubalBeltran
2,8061916
HINT: divide the ecuation by $costheta$
answered Jul 14 '14 at 1:52
AsdrubalBeltran
2,8061916
answered Jul 14 '14 at 1:52
AsdrubalBeltran
2,8061916
answered Jul 14 '14 at 1:52
AsdrubalBeltran
2,8061916
answered Jul 14 '14 at 1:52
AsdrubalBeltran
2,8061916
2,8061916
add a comment |
add a comment |
up vote
0
down vote
Divide both sides by 5, and observe that the coefficients satisfy
$left(frac{3}{5}right)^2+ left(frac{4}{5}right)^2=1.$ Consequently there is some angle $alpha$ such that $cos alpha = 3/5, sinalpha =4/5$. (For comparison with other answers, note that $tanalpha = 4/3$). With this in mind we can use the sum-to-product identity to write
$$ sin(alpha+theta) = cosalpha sin theta+sinalpha costheta = frac{3}{5}sintheta+frac{4}{5}costheta = 0.$$ So all that's left is to find the solutions of this equation.
answered Jul 14 '14 at 2:08
Semiclassical
11k32464
add a comment |
up vote
0
down vote
Divide both sides by 5, and observe that the coefficients satisfy
$left(frac{3}{5}right)^2+ left(frac{4}{5}right)^2=1.$ Consequently there is some angle $alpha$ such that $cos alpha = 3/5, sinalpha =4/5$. (For comparison with other answers, note that $tanalpha = 4/3$). With this in mind we can use the sum-to-product identity to write
$$ sin(alpha+theta) = cosalpha sin theta+sinalpha costheta = frac{3}{5}sintheta+frac{4}{5}costheta = 0.$$ So all that's left is to find the solutions of this equation.
answered Jul 14 '14 at 2:08
Semiclassical
11k32464
add a comment |
up vote
0
down vote
up vote
0
down vote
Divide both sides by 5, and observe that the coefficients satisfy
$left(frac{3}{5}right)^2+ left(frac{4}{5}right)^2=1.$ Consequently there is some angle $alpha$ such that $cos alpha = 3/5, sinalpha =4/5$. (For comparison with other answers, note that $tanalpha = 4/3$). With this in mind we can use the sum-to-product identity to write
$$ sin(alpha+theta) = cosalpha sin theta+sinalpha costheta = frac{3}{5}sintheta+frac{4}{5}costheta = 0.$$ So all that's left is to find the solutions of this equation.
answered Jul 14 '14 at 2:08
Semiclassical
11k32464
Divide both sides by 5, and observe that the coefficients satisfy
$left(frac{3}{5}right)^2+ left(frac{4}{5}right)^2=1.$ Consequently there is some angle $alpha$ such that $cos alpha = 3/5, sinalpha =4/5$. (For comparison with other answers, note that $tanalpha = 4/3$). With this in mind we can use the sum-to-product identity to write
$$ sin(alpha+theta) = cosalpha sin theta+sinalpha costheta = frac{3}{5}sintheta+frac{4}{5}costheta = 0.$$ So all that's left is to find the solutions of this equation.
answered Jul 14 '14 at 2:08
Semiclassical
11k32464
answered Jul 14 '14 at 2:08
Semiclassical
11k32464
answered Jul 14 '14 at 2:08
Semiclassical
11k32464
answered Jul 14 '14 at 2:08
Semiclassical
11k32464
11k32464
add a comment |
add a comment |
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Hint: convert to a ratio relationship.
– Gina
Jul 14 '14 at 1:46
@Gina Why don't you make that an answer? (Maybe be a little more explicit.)
– Cameron Williams
Jul 14 '14 at 1:50
Which do you use 360 (rad), or 360 (degree)?
– choco_addicted
Nov 29 at 6:26