trignometry?

Question

i dont understand how you solve

2cos^2feta+3sinfeta=3

please explain your wrking out
nd does the method describe appliy to all trig equations

Answer 1

Do you mean "theta" instead of feta?

2 cos² Θ + 3 sin Θ = 3 would be your equation in that case. I think cos² Θ = (1 - sin² Θ), the substitution of which would work to give you a quadratic equation.

Answer 2

You refer to an angle "feta".
It should actually be called theta.
I don`t know how to get theta on my keyboard so, if you don`t mind , I will call it X!
The equation then becomes :-
2cos²X+3sinX = 3
2(1 - sin²X) + 3sinX = 3
2 - 2sin²X + 3sinX = 3
2sin²X - 3sinX +1= 0
(2sinX - 1)(sinx - 1) =0
sinX=1/2 , sinX=1
X=30°,150°,90°

Answer 3

2cos^2feta+3sinfeta=3
cos^2 = 1-sin^2 so
2(1-sin^2 t)+3sin t=3
-2sin^2 t +3sin t +2=3
-2sin^2 t+3sin t-1=0
2sin^2 t-3sin t+1=0
(2sin t -1)(sin t -1)=0
2 sin t-1=0
sin t=1/2
t=arcsin 1/2
t=π/6 or 30°
sin t-1=0
sin t=1
t=arcsin 1
t=π/2 or 90°
theta can be 30°, 90° or 150°

Answer 4

2cos^2x +3sin x -3 =0, where x = feta
Use cos^2x = 1-sin^2x to get
2(1-sin^2x) +3sinx -3 = 0
2-2sin^2 x +3 sin x -3 = 0
-2sin^2x +3sinx - 1 = 0
Let y=sinx
So -2y^2 +3y -1 =0
(-2y+1)(y-1) =0
y = 1 or y = 1/2
So sin x = 1 or sin x =1/2
So x = 90 degrees or x = 30 degrees

Answer 5

Do your own homework or pay me to tutor you!

Answer 6

Do you mean theta??