3-2(Cosx)^2 - 3Sinx = 0.

Recall (Sinx)^2 + (Cosx)^2 = 1.
Therefore (Cosx)^2 = 1 - (Sinx)^2
Substitute this into the question above.

3-2(Cosx)^2 - 3Sinx = 0
3 - 2(1 - (Sinx)^2) - 3Sinx = 0 Expand
3 - 2 + 2(Sinx)^2 - 3Sinx = 0
1 + 2(Sinx)^2 - 3Sinx = 0 Rearrange
2(Sinx)^2 - 3Sinx + 1 = 0
Let p = Sinx
2p^2 - 3p + 1 = 0 Factorise the quadratic expression
2p^2 - p - 2p +1 = 0
p(2p -1) - 1(2p -1) = 0
(2p-1)(p -1) = 0

Therefore 2p-1=0 or (p-1) = 0
2p=0+1 or (p-1) = 0
2p=1 or p = 0 +1.
p=1/2 or p = 1 Recall p = Sinx

Therefore Sinx = 1/2 or 1.
For 0<x<360

Sinx =1/2, x = Sin inverse (1/2) , x = 30,
 (180-30)- 2nd Quadrant = 150 deg
Sinx = 1, x = Sin inverse (1) , x = 90

Therefore x = 30,90 & 150 degrees.

Cheers.

7 0

3 years ago

(7 + (-3) - 4) ÷ ( ( -7) x ( 2- (-6) ) ) Please show work!

kotegsom [21]

Answer:

The answer is attached to the picture