Solve On The Interval \[0, 2\pi$\]:$(\sin X - 1)(2 \sin^2 X - 5 \sin X + 2) = 0$A. $x = \frac{\pi}{2}, X = \frac{\pi}{6}, X = \frac{5\pi}{6}$ B. $x = \pi, X = \frac{2\pi}{3}, X = \frac{5\pi}{3}$ C. $x = 2\pi, X

Feb 25, 2025 by ADMIN 213 views

Solve on the Interval [0, 2π]: (sin x - 1)(2 sin^2 x - 5 sin x + 2) = 0

Introduction

In this article, we will be solving a trigonometric equation on the interval [0, 2π]. The given equation is (sin x - 1)(2 sin^2 x - 5 sin x + 2) = 0. We will use various trigonometric identities and techniques to solve this equation and find the values of x that satisfy it.

Step 1: Factor the Quadratic Expression

The given equation can be factored as (sin x - 1)(2 sin^2 x - 5 sin x + 2) = 0. We can factor the quadratic expression 2 sin^2 x - 5 sin x + 2 as (2 sin x - 1)(sin x - 2) = 0.

Step 2: Set Each Factor Equal to Zero

To find the values of x that satisfy the equation, we need to set each factor equal to zero and solve for x.

2.1: sin x - 1 = 0

We can solve the equation sin x - 1 = 0 by adding 1 to both sides, which gives us sin x = 1. The sine function has a value of 1 at x = π/2 and x = 5π/6.

2.2: 2 sin x - 1 = 0

We can solve the equation 2 sin x - 1 = 0 by adding 1 to both sides and then dividing by 2, which gives us sin x = 1/2. The sine function has a value of 1/2 at x = π/6 and x = 5π/6.

2.3: sin x - 2 = 0

We can solve the equation sin x - 2 = 0 by adding 2 to both sides, which gives us sin x = 2. However, the sine function has a maximum value of 1, so there are no solutions to this equation.

Step 3: Find the Values of x

We have found the values of x that satisfy each factor of the equation. Now, we need to find the values of x that satisfy the original equation.

3.1: x = π/2

We have found that sin x = 1 at x = π/2. Therefore, x = π/2 is a solution to the original equation.

3.2: x = π/6

We have found that sin x = 1/2 at x = π/6. Therefore, x = π/6 is a solution to the original equation.

3.3: x = 5π/6

We have found that sin x = 1/2 at x = 5π/6. Therefore, x = 5π/6 is a solution to the original equation.

Conclusion

In this article, we have solved the trigonometric equation (sin x - 1)(2 sin^2 x - 5 sin x + 2) = 0 on the interval [0, 2π]. We have found that the values of x that satisfy the equation are x = π/2, x = π/6, and x = 5π/6.

Discussion

The given equation is a quadratic equation in terms of sin x. We have factored the quadratic expression and set each factor equal to zero to find the values of x that satisfy the equation. We have also used the properties of the sine function to find the values of x that satisfy the equation.

Final Answer

The final answer is x = π/2, x = π/6, and x = 5π/6.

References

[1] "Trigonometry" by Michael Corral
[2] "Calculus" by Michael Spivak