Determine the general solution for the following: 1.1 sin2 x - sin x = cos2 x 1.2 sin2 x + sin x cos x = 0

Here are the solutions to your trigonometry problems: 1.1 Determine the general solution for ^2 x - x = ^2 x Step 1: Use the identity ^2 x = 1 - ^2 x to rewrite the equation in terms of x. ^2 x - x = 1 - ^2 x Step 2: Rearrange the equation into a quadratic form. 2^2 x - x - 1 = 0 Step 3: Let y = x. Solve the quadratic equation 2y^2 - y - 1 = 0. Factoring the quadratic: (2y + 1)(y - 1) = 0 This gives two possible values for y: 2y + 1 = 0 y = -(1)/(2) y - 1 = 0 y = 1 So, x = -(1)/(2) or x = 1. Step 4: Find the general solutions for x. For x = 1: The principal value is x = 90^. The general solution is x = 90^ + 360^ k, where k Z. For x = -(1)/(2): The reference angle is 30^. Since x is negative, x lies in the third or fourth quadrants. In the third quadrant: x = 180^ + 30^ = 210^. In the fourth quadrant: x = 360^ - 30^ = 330^. The general solutions are x = 210^ + 360^ k and x = 330^ + 360^ k, where k Z. The general solution is x = 90^ + 360^ k, x = 210^ + 360^ k, x = 330^ + 360^ k, k Z. 1.2 Determine the general solution for ^2 x + x x = 0 Step 1: Factor out x. x ( x + x) = 0 Step 2: Set each factor to zero and solve for x. Case 1: x = 0 The general solution is x = 180^ k, where k Z. Case 2: x + x = 0 Divide by x (assuming x ≠ 0. If x = 0, then x = ± 1, so x + x = ± 1 ≠ 0. Thus x ≠ 0 is a valid assumption). ( x)/( x) + ( x)/( x) = 0 x + 1 = 0 x = -1 The principal value for x = -1 is x = -45^ or x = 135^. The general solution is x = 135^ + 180^ k, where k Z. The general solution is x = 180^ k, x = 135^ + 180^ k, k Z. 2. Determine the general solution of 3 2x = 1 + 5 x Step 1: Use the double angle identity 2x = 2^2 x - 1 to rewrite the equation in terms of x. 3(2^2 x - 1) = 1 + 5 x Step 2: Expand and rearrange the equation into a quadratic form. 6^2 x - 3 = 1 + 5 x 6^2 x - 5 x - 4 = 0 Step 3: Let y = x. Solve the quadratic equation 6y^2 - 5y - 4 = 0. Using the quadratic formula y = -b ± sqrt(b^2 - 4ac)2a: y = -(-5) ± sqrt((-5)^2 - 4(6)(-4))2(6) y = 5 ± sqrt(25 + 96)12 y = 5 ± sqrt(121)12 y = (5 ± 11)/(12) This gives two possible values for y: y_1 = (5 + 11)/(12) = (16)/(12) = (4)/(3) y_2 = (5 - 11)/(12) = (-6)/(12) = -(1)/(2) So, x = (4)/(3) or x = -(1)/(2). Step 4: Find the general solutions for x. For x = (4)/(3): This has no solution because the range of x is [-1, 1]. For x = -(1)/(2): The principal value is x = 120^. The general solution is x = ± 120^ + 360^ k, where k Z. The general solution is x = ± 120^ + 360^ k, k Z. 3. Solve for x: 4^2 x - 3 x - 1 = 0 for x [0^; 360^] Step 1: This is a quadratic equation in terms of x. Let y = x. 4y^2 - 3y - 1 = 0 Step 2: Solve the quadratic equation for y. Factoring the quadratic: (4y + 1)(y - 1) = 0 This gives two possible values for y: 4y + 1 = 0 y = -(1)/(4) y - 1 = 0 y = 1 So, x = -(1)/(4) or x = 1. Step 3: Find the values of x in the range [0^; 360^]. For x = 1: x = 90^ For x = -(1)/(4): The reference angle is = ((1)/(4)) ≈ 14.48^. Since x is negative, x lies in the third or fourth quadrants. In the third quadrant: x = 180^ + ≈ 180^ + 14.48^ = 194.48^. In the fourth quadrant: x = 360^ - ≈ 360^ - 14.48^ = 345.52^. The values of x in the given range are x = 90^, 194.48^, 345.52^. 4. Determine the general solution of x = 2 2x where x < 0. Step 1: Rewrite x as ( x)/( x) and 2x as 2 x x. ( x)/( x) = 2(2 x x) ( x)/( x) = 4 x x Step 2: Rearrange the equation to solve for x. x = 4 x ^2 x x - 4 x ^2 x = 0 x (1 - 4^2 x) = 0 Step 3: Set each factor to zero. Case 1: x = 0 The general solution is x = 180^ k, where k Z. We also need to consider the condition x < 0. If x = 180^ k: If k is even, x = 0^, 360^, , then x = 1. This does not satisfy x < 0. If k is odd, x = 180^, 540^, , then x = -1. This satisfies x < 0. So, for this case, x = 180^ (2m+1), where m Z. This can be written as x = 180^ + 360^ m. Case 2: 1 - 4^2 x = 0 4^2 x = 1 ^2 x = (1)/(4) x = ± (1)/(2) We are given the condition x < 0, so we must have x = -(1)/(2). For x = -(1)/(2): The principal value is x = 120^. The general solution is x = ± 120^ + 360^ k, where k Z. Both 120^ and -120^ (or 240^) have negative cosine values. So, x = 120^ + 360^ k and x = 240^ + 360^ k (or x = -120^ + 360^ k). Combining the valid solutions from both cases: The general solution is x = 180^ + 360^ k, x = 120^ + 360^ k, x = 240^ + 360^ k, k Z. 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 quadratic form. 2^2 x - x - 3 = 0 Step 3: Let y = x. Solve the quadratic equation 2y^2 - y - 3 = 0. Factoring the quadratic: (2y - 3)(y + 1) = 0 This gives two possible values for y: 2y - 3 = 0 y = (3)/(2) y + 1 = 0 y = -1 So, x = (3)/(2) or x = -1. Step 4: Find the values of x in the range [-180^; 360^]. For x = (3)/(2): This has no solution because the range of x is [-1, 1]. For x = -1: The values of x in the range [-180^; 360^] are: x = 180^ x = -180^ The values of x in the given range are x = -180^, 180^. What's next? Send 'em! 📸