Quadratic Trigonometric Equations

Solving trig equations using identities

0:00 Intro
0:23 Simple example
5:50 Example 2
6:52 Identity
16:30 Example 3
24:05 tan question

[jan-willemreens]

...Good day Miss Jay, I hope your doing well. First of all thank you for your nice presentation on Quadratic Trigonometric Equations. Last problem (let Theta = T) : (-360deg </= T </= 360deg), 3(tanT)^2 = 1 <---> (tanT)^2 = 1/3 <---> tanT = (1/3)SQRT(3) or tanT = (-1/3)SQRT(3) ---> T = {-330deg, -210deg, -150deg, -30deg, 30deg, 150deg, 210deg, 330deg}, so a total of 8 solutions. Miss Jay, I hope for a positive response, and thank you again for your pleasant live stream... Jan-W

[jan-willemreens]

...Miss Jay, I forgot to make a small remark when looking for and listing the solutions of an equation. In addition to sketching a graph, I personally also like to draw a unit circle, so that you can see in which quadrants solutions may lie. Of course, each individual has her/his own preference for solving such equations. When dealing with Trigonometric problems, the first thing that comes to my mind is the good old trustworthy unit circle! Jan-W

[geraldillo]

Thanks for posting!
here is my attempt to solve;

3tan^2 θ=1
1=3tan^2 θ
1=3(sin^2 θ/cos^2 θ)
1=(3sin^2 θ)/cos^2 θ
cos^2 θ =3sin^2 θ
cos^2 θ =3(1-cos^2 θ)
cos^2 θ =3-3cos^2 θ
4cos^2 θ=3
cos^2 θ=3/4

cosθ=√3/2
cosθ=±210º or ±150º

cosθ= -√3/2
cosθ=±30º or ±330º