Determine all solutions in the interval
x is all real numbers , [0, 2 pi]
using a trigonometric identify
2cos^2x + sinx - 1 = 0
2cos^2x + sinx - 1 = 0
2(1 - sin^2x) + sinx - 1 = 0
2 - 2sin^2x + sinx - 1 = -
2sin^2x - sinx -1 = 0
(2sinx + 1)(sinx - 1) = 0
sinx = -1/2 or sinx = 1
from sinx = -1/2, x = pi + pi/6 or 2pi - pi/6
x = 7pi/6 radians or 11pi/6 radians ,
(210º or 330º)
from sinx = 1, x = pi/4 radians, (90º)
To solve the equation 2cos^2x + sinx - 1 = 0 using a trigonometric identity, we can use the Pythagorean identity for cosine. The Pythagorean identity states that cos^2x + sin^2x = 1.
Let's start by rearranging the equation to include the cosine identity:
2cos^2x + sinx - 1 = 0
Using the Pythagorean identity, substitute cos^2x with 1 - sin^2x:
2(1 - sin^2x) + sinx - 1 = 0
Simplify the equation:
2 - 2sin^2x + sinx - 1 = 0
Combine like terms:
-2sin^2x + sinx + 1 = 0
Now, let's factor the quadratic equation. Notice that the equation is in the form of as^2 + bs + c = 0, where a = -2, b = 1, and c = 1.
To factor the equation, we need to find two numbers (m and n) whose product is equal to the product of a and c, and whose sum is equal to b.
The product of -2 and 1 is -2. The sum of -2 and 1 is -1.
Therefore, we can rewrite the equation as:
-2sin^2x - sinx + 1 = 0
Next, we factor the quadratic expression:
(-2sinx + 1)(sinx + 1) = 0
Now, set each factor equal to zero and solve for x:
-2sinx + 1 = 0 or sinx + 1 = 0
For -2sinx + 1 = 0:
-2sinx = -1
sinx = 1/2
Using the interval [0, 2π], the solutions for sinx = 1/2 are x = π/6 and x = 5π/6.
For sinx + 1 = 0:
sinx = -1
This occurs at x = 3π/2.
Therefore, the solutions to the equation 2cos^2x + sinx - 1 = 0 in the interval [0, 2π] are x = π/6, 5π/6, and 3π/2.