QuestionJuly 14, 2025

Question 2 Solve the equation: sin(2x)-cos(2x)=1 over the interval [0,2pi )

Question 2 Solve the equation: sin(2x)-cos(2x)=1 over the interval [0,2pi )
Question 2
Solve the equation: sin(2x)-cos(2x)=1 over the interval [0,2pi )

Solution
4.5(238 votes)

Answer

x = \frac{\pi}{4}, \frac{5\pi}{4} Explanation 1. Use Trigonometric Identity Apply the identity \sin(2x) = 2\sin(x)\cos(x) and \cos(2x) = \cos^2(x) - \sin^2(x). 2. Simplify Equation Substitute identities into equation: 2\sin(x)\cos(x) - (\cos^2(x) - \sin^2(x)) = 1. 3. Rearrange Terms Rearrange to form a quadratic in terms of \sin(x) or \cos(x). 4. Solve Quadratic Equation Solve for \sin(x) or \cos(x). Check solutions within interval [0, 2\pi). 5. Verify Solutions Ensure solutions satisfy original equation.

Explanation

1. Use Trigonometric Identity<br /> Apply the identity $\sin(2x) = 2\sin(x)\cos(x)$ and $\cos(2x) = \cos^2(x) - \sin^2(x)$.<br />2. Simplify Equation<br /> Substitute identities into equation: $2\sin(x)\cos(x) - (\cos^2(x) - \sin^2(x)) = 1$.<br />3. Rearrange Terms<br /> Rearrange to form a quadratic in terms of $\sin(x)$ or $\cos(x)$.<br />4. Solve Quadratic Equation<br /> Solve for $\sin(x)$ or $\cos(x)$. Check solutions within interval $[0, 2\pi)$.<br />5. Verify Solutions<br /> Ensure solutions satisfy original equation.
Click to rate:

Similar Questions