2cos^2x-3sinx=0
2(1 - sin^2x) - 3sinx = 0
-2sin^2x - 3sinx + 2 = 0
2sin^2x + 3sinx - 2 = 0
(2sinx -1)(sinx + 2) = 0
sinx = 1/2 or sinx = -2
x = 30° or 150° or (sinx = -2 has no solution)
x = 30 or 150 ° which is also π/6 or 5π/6 radians
Thanks
To solve the equation 2cos^2(x) - 3sin(x) = 0, we can use basic trigonometric identities and algebraic techniques. Here's how you can go about it:
Step 1: Use the Pythagorean identity for cos^2(x) and sin^2(x): sin^2(x) + cos^2(x) = 1.
Step 2: Use the given equation to substitute cos^2(x) in terms of sin(x): cos^2(x) = 1 - sin^2(x).
Step 3: Substitute cos^2(x) in the equation:
2(1 - sin^2(x)) - 3sin(x) = 0.
Step 4: Expand and simplify the equation:
2 - 2sin^2(x) - 3sin(x) = 0.
Step 5: Rearrange the terms to form a quadratic equation:
-2sin^2(x) - 3sin(x) + 2 = 0.
Step 6: Solve the quadratic equation. Since the equation is in terms of sin(x), let's make a substitution: let y = sin(x).
Step 7: Substituting y in the equation: -2y^2 - 3y + 2 = 0.
Step 8: Factor the quadratic equation: (-2y + 1)(y + 2) = 0.
Step 9: Set each factor equal to zero and solve for y:
-2y + 1 = 0 or y + 2 = 0.
Step 10: Solve for y:
-2y = -1 or y = -2.
Step 11: Solve for sin(x) using the values of y:
-2y = -1: -2sin(x) = -1, so sin(x) = 1/2.
y = -2: sin(x) = -2 (not valid, as sin(x) should be between -1 and 1).
Step 12: Solve for the values of x using the value of sin(x):
sin(x) = 1/2 can occur in two quadrants: the first quadrant (0 < x < π/2) and the second quadrant (π/2 < x < π).
Step 13: Find the solutions for x in the first quadrant:
sin(x) = 1/2: x = π/6.
Step 14: Find the solutions for x in the second quadrant:
sin(x) = 1/2: x = π - π/6 = 5π/6.
Therefore, the solutions to the equation 2cos^2(x) - 3sin(x) = 0 are x = π/6 and x = 5π/6.