Answer:
{π/4, 5π/4}
Step-by-step explanation:
Tan theta -1=0 could be rewritten as tan Ф = 1. The tangent function is 1 at Ф = π/4. As the period of the tangent function is π,
tan Ф = 1 will be true for Ф = π/4 + π, or (5/4)π.
The solution set is {π/4, 5π/4}.
Sub x = 2-y^2 to Q, we get:
Q = 3(2-y^2)*y^2
let y^2 = k
Q = 3(2-k)k = 3(2k-k^2)
2k-k^2 has a max when k = 1
Then y^2 = 1 -> y = 1 or -1
Answer:im pretty sure its 2 then you can watch an ad for another answer
Step-by-step explanation:
Answer: I don't know what you wanted to be solved, but, I solved for x
Step-by-step explanation:
<u>Solved for x</u>
- <u>x=(2\pm i\sqrt(6))/(2)</u>
