The problem presents a trigonometric equation that can be solved by treating it as a quadratic equation. We will solve the equation:

Hey Balogun, good to see you again. Step 1: Identify the equation to solve. The problem presents a trigonometric equation that can be solved by treating it as a quadratic equation. We will solve the equation: 2 ^2 - - 1 = 0 Step 2: Substitute a variable to form a quadratic equation. Let x = . Substituting this into the equation gives a standard quadratic form: 2x^2 - x - 1 = 0 Step 3: Solve the quadratic equation for x. We can factor the quadratic equation. We look for two numbers that multiply to (2)(-1) = -2 and add up to -1. These numbers are -2 and 1. 2x^2 - 2x + x - 1 = 0 Factor by grouping: 2x(x - 1) + 1(x - 1) = 0 (2x + 1)(x - 1) = 0 This yields two possible values for x: 2x + 1 = 0 2x = -1 x = -(1)/(2) x - 1 = 0 x = 1 Step 4: Substitute back and solve for . Now we substitute back for x: Case 1: = 1 The general solution for = 1 is when is ()/(2) plus any multiple of 2. = ()/(2) + 2n where n is an integer. Case 2: = -(1)/(2) The reference angle for = (1)/(2) is = ()/(6) (30^). Since is negative, must be in the third or fourth quadrants. In the third quadrant: = + ()/(6) = (6)/(6) + ()/(6) = (7)/(6) In the fourth quadrant: = 2 - ()/(6) = (12)/(6) - ()/(6) = (11)/(6) The general solutions for = -(1)/(2) are: = (7)/(6) + 2n = (11)/(6) + 2n where n is an integer. Step 5: State the complete set of general solutions. Combining all possible values for , the general solutions are: = ()/(2) + 2n, = (7)/(6) + 2n, = (11)/(6) + 2n where n Z What's next? Send 'em! 📸