Asked by lindsay
Find all the angles between 0 degrees and 360 degrees which satisfy the equations
a. 8 cos x sin x = sin x
b. 5 tan^2 y + 5 tan y = 2 sec^2 y.
a. 8 cos x sin x = sin x
b. 5 tan^2 y + 5 tan y = 2 sec^2 y.
Answers
Answered by
Reiny
a)
8cosxsinx - sinx = 0
sinx(8cosx - 1) = 0
sinx = 0 or cosx = 1/8
if sinx = 0 , x = 0, 180, 360
if cosx = 1/8, x = 82.8° or 277.18°
working on the 2nd....
8cosxsinx - sinx = 0
sinx(8cosx - 1) = 0
sinx = 0 or cosx = 1/8
if sinx = 0 , x = 0, 180, 360
if cosx = 1/8, x = 82.8° or 277.18°
working on the 2nd....
Answered by
Reiny
b)
5 tan^2 y + 5 tan y = 2 sec^2 y
playing around with some right-angled triangles, using the numbers 2 and 5
I drew a triangle with base 1, height 2 and hypotenuse 5
let the base angle be y
then tany = 2/1 , y = 63.4349...° ( stored in memory)
checking:
LS = 5(2/1)^2 + 5(2/1) = 30
RS = 2/(1/√5)^2 = 10
ahh, but if tany = -2/1
LS = 5(-2/1)^2 + 5(-2/1) = 10
RS = 1/(1/√5)^2 = 10
so we need y = 180-63.4349 or y = 360-63.4349
y = 116.565° or y = 296.565°
5 tan^2 y + 5 tan y = 2 sec^2 y
playing around with some right-angled triangles, using the numbers 2 and 5
I drew a triangle with base 1, height 2 and hypotenuse 5
let the base angle be y
then tany = 2/1 , y = 63.4349...° ( stored in memory)
checking:
LS = 5(2/1)^2 + 5(2/1) = 30
RS = 2/(1/√5)^2 = 10
ahh, but if tany = -2/1
LS = 5(-2/1)^2 + 5(-2/1) = 10
RS = 1/(1/√5)^2 = 10
so we need y = 180-63.4349 or y = 360-63.4349
y = 116.565° or y = 296.565°
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