Calculate the value of x, if x ∈ [-180°; 360°] cos2x = cosx + 2

Here's the solution for question 5: 5. Calculate the value of x, if x [-180^; 360^] for 2x = x + 2 Step 1: Use the double angle identity 2x = 2^2 x - 1 to rewrite the equation in terms of x. 2^2 x - 1 = x + 2 Step 2: Rearrange the equation into a standard quadratic form. 2^2 x - x - 1 - 2 = 0 2^2 x - x - 3 = 0 Step 3: Let y = x. Solve the quadratic equation 2y^2 - y - 3 = 0. Factor the quadratic expression: (2y - 3)(y + 1) = 0 This gives two possible values for y: 2y - 3 = 0 y = (3)/(2) y + 1 = 0 y = -1 So, we have x = (3)/(2) or x = -1. Step 4: Determine the values of x in the given range [-180^; 360^]. For x = (3)/(2): There is no solution for this, as the range of the cosine function is [-1, 1]. The value (3)/(2) = 1.5 is outside this range. For x = -1: The values of x for which x = -1 are 180^, -180^, 540^, . Considering the specified interval x [-180^; 360^], the values are: x = -180^ x = 180^ The values of x in the given range are x = -180^, 180^. 3 done, 2 left today. You're making progress.