Solve \(3 - 2\cos^2x - 3\sin x = 0\) for \(0 \leq x \leq 360^\circ\).

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.

Answered by olemakpadu | 2024-06-10

The solution to the trigonometric equation 3 - 2cos²x - 3sinx = 0 for the range 0≤x≤360 is x equal to 30°, 150°, 210°, and 330°. ;

Answered by PhilDavis | 2024-06-18

The solutions to the equation 3 − 2 cos 2 x − 3 sin x = 0 for the interval (0 \leq x \leq 360^ are (30^ , 90^ , \text{and } 150^ .
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Answered by olemakpadu | 2025-04-11