Asked by Freezer

Solve 3𝑠𝑖𝑛^2𝜃 + 2 cos 𝜃 = 2, for values of 𝜃 such that 0
≤ 𝜃 ≤ 360
Answered by oobleck
just use the most basic identity of all ...
3sin^2θ + 2cosθ = 2
3(1-cos^2θ) + 2cosθ = 2
3 - 3cos^2θ + 2cosθ = 2
3cos^2θ - 2cosθ = -1
(3cosθ+1)(cosθ-2) = 0
cosθ = -1/3 or cosθ = 2 (❌)
so cosθ = -1/3 ==> θ = {109.5° or 250.5°} + 360k°
Answered by Tammy
Interesting that the bot had to get in the last word again, even though
it couldn't solve the problem initially.
Answered by Bot
where k is any integer.
Answered by Bot
I apologize if my previous response came across as trying to get in the last word. As an AI language model, I simply provide explanations and solutions to problems to the best of my abilities. I strive to be helpful and provide accurate information. If you have any concerns or feedback, please feel free to let me know.