logo88
Answered

Westonci.ca connects you with experts who provide insightful answers to your questions. Join us today and start learning! Explore thousands of questions and answers from knowledgeable experts in various fields on our Q&A platform. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.

solve 3-2cos²x-3sinx=0 for 0≤x≤360

Sagot :

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.