answer:To solve the equation 2sin(2x) - 1 = 0 in the interval [0, 2π), we first isolate sin(2x): 2sin(2x) = 1 sin(2x) = 1/2 Now, we find the angles whose sine is 1/2. Since the sine function is positive in the first and second quadrants, we can find the principal angles in those quadrants: sin(π/6) = 1/2 sin(5π/6) = 1/2 Since sin(2x) = 1/2, we have two cases: Case 1: 2x = π/6 x = π/12 Case 2: 2x = 5π/6 x = 5π/12 However, we also need to consider the periodicity of the sine function. The sine function has a period of 2π, so we can add 2πn to the angles, where n is an integer, and still get the same sine value. For our equation, we have: 2x = π/6 + 2πn x = π/12 + πn and 2x = 5π/6 + 2πn x = 5π/12 + πn Now, we need to find the values of n that give us solutions in the interval [0, 2π): For x = π/12 + πn: n = 0: x = π/12 n = 1: x = 13π/12 n = 2: x = 25π/12 (not in the interval [0, 2π)) For x = 5π/12 + πn: n = 0: x = 5π/12 n = 1: x = 17π/12 n = 2: x = 29π/12 (not in the interval [0, 2π)) So, the solutions in the interval [0, 2π) are: x = π/12, 5π/12, 13π/12, and 17π/12

answer:To find all solutions of the equation 2cos(3x) = √3 in the interval [0, 2π], follow these steps: 1. Isolate the cosine function: cos(3x) = (√3)/2 2. Find the general solutions for x: Since cos(π/6) = cos(5π/6) = (√3)/2, we have two solutions for the cosine function in the interval [0, 2π]: 3x = π/6 + 2nπ and 3x = 5π/6 + 2nπ, where n is an integer. 3. Solve for x: x = (π/18) + (2nπ/3) and x = (5π/18) + (2nπ/3) 4. Find all specific solutions in the interval [0, 2π]: For the first equation, x = (π/18) + (2nπ/3), we have: n = 0: x = π/18 n = 1: x = π/18 + 2π/3 = 13π/18 n = 2: x = π/18 + 4π/3 = 25π/18 > 2π (out of the interval) For the second equation, x = (5π/18) + (2nπ/3), we have: n = 0: x = 5π/18 n = 1: x = 5π/18 + 2π/3 = 17π/18 n = 2: x = 5π/18 + 4π/3 = 29π/18 > 2π (out of the interval) So, the solutions of the equation 2cos(3x) = √3 in the interval [0, 2π] are x = π/18, 13π/18, 5π/18, and 17π/18.

answer:To solve the equation sin(2x) + cos(x) = 0, we can use the double-angle identity for sine, which states that sin(2x) = 2sin(x)cos(x). So, the equation becomes: 2sin(x)cos(x) + cos(x) = 0 Now, we can factor out cos(x): cos(x)(2sin(x) + 1) = 0 This equation is satisfied if either of the factors is equal to zero. So, we have two separate cases to solve: 1) cos(x) = 0 2) 2sin(x) + 1 = 0 For case 1, cos(x) = 0: x = 90° and x = 270° (since cosine is zero at these angles within the given range of 0 to 360 degrees) For case 2, 2sin(x) + 1 = 0: sin(x) = -1/2 x = arcsin(-1/2) x ≈ -30° and x ≈ -150° (in the range of -180° to 180°) However, we need to find the solutions in the range of 0 to 360 degrees. To do this, we can add 360° to each of the negative angles: x ≈ 330° and x ≈ 210° So, the solutions for x within the range of 0 to 360 degrees are: x = 90°, 270°, 330°, and 210°.

answer:To find the value of theta (in degrees) for the equation cos(theta) = 0.5, we need to find the angle whose cosine is 0.5. The cosine function has a value of 0.5 at two angles in the range of 0 to 360 degrees: 1. Theta = 60 degrees (This is because cos(60°) = 0.5) 2. Theta = 300 degrees (This is because cos(300°) = 0.5) So, the two possible values of theta (in degrees) for the equation cos(theta) = 0.5 are 60 degrees and 300 degrees.