Question

2 sin^2 x +3 cos x - 3 = 0

Answer 1

$\displaystyle 2sin^2x+3cosx-3=0\\2(1-cos^2x)+3cosx-3=0\\2-2cos^2x+3cosx-3=0\\2cos^2x-3cosx+1=0\\(cosx)_{1,2}=\frac{3\pm\sqrt{9-8}}{4}=\frac{3\pm1}{4}\\cosx_1=1\ \ \ \ \ \ \ \ \ \ \ \ cosx_2=\frac{1}{2}\\x_1=2\pi n;\ \ \ \ \ \ \ \ \ \ \ x_2=\pm\frac{\pi}{3}+2\pi n;n\in Z$

Answer 2

The equation you've provided is a trigonometric equation involving both sine and cosine functions. To solve for the values of x that satisfy the equation, we can follow these steps:

1. Use the trigonometric identity: sin^2(x) + cos^2(x) = 1.
2. Express sin^2(x) in terms of cos(x) using the identity: sin^2(x) = 1 - cos^2(x).
3. Substitute the expression for sin^2(x) into the equation.

Let's solve it step by step:

Given equation: 2sin^2(x) + 3cos(x) - 3 = 0

Step 1: sin^2(x) = 1 - cos^2(x)

Step 2: 2(1 - cos^2(x)) + 3cos(x) - 3 = 0 2 - 2cos^2(x) + 3cos(x) - 3 = 0

Step 3: -2cos^2(x) + 3cos(x) - 1 = 0

Now we have a quadratic equation in terms of cos(x). To solve for cos(x), we can factor or use the quadratic formula:

-2cos^2(x) + 3cos(x) - 1 = 0

Let's use the quadratic formula: cos(x) = (-B ± √(B^2 - 4AC)) / 2A

Where A = -2, B = 3, and C = -1.

cos(x) = (-3 ± √(3^2 - 4*(-2)(-1))) / (2(-2)) cos(x) = (-3 ± √(9 - 8)) / (-4) cos(x) = (-3 ± √1) / (-4) cos(x) = (-3 ± 1) / (-4)

This gives two possible solutions for cos(x):

1. cos(x) = -1
2. cos(x) = -1/2

Now, we can find the corresponding values of x for these solutions:

1. For cos(x) = -1: x = π

2. For cos(x) = -1/2: Since cosine is negative in the 2nd and 3rd quadrants, we have: x = 2π/3 (2nd quadrant) x = 4π/3 (3rd quadrant)

So, the solutions for the equation 2sin^2(x) + 3cos(x) - 3 = 0 are: x = π, 2π/3, 4π/3