Solve for x, between 0 and 360 degrees, 4cos2 (x) + 7sin (x) – 2 = 0

Solve for x, between 0 and 360 degrees, 4cos² (x) + 7sin (x) – 2 = 0 The way to approach this problem is to understand the relation that: sin² (x) + cos² (x) = 1 That equation can be rearranged to give: cos² (x) = 1 - sin² (x) Substituting that into the question, we get: 4 (1 - sin² (x)) + 7sin (x) – 2 = 0 4 - 4sin² (x) + 7sin (x) – 2 = 0 Rearranging and simplifing the above gives: 4sin² (x) - 7sin (x) + 2 = 0 Now that is beginning to look like a quadratic! Substituting sin (x) with y (i.e. y = sin (x)), you get: 4y² - 7y + 2 = 0 Using the quadratic equation the roots can be calculated: y = 2 or y = -0.25 Remembering that y = sin (x) sin (x) = 2 or sin (x) = -0.2 Since the range of possible values for sin (x) are between -1 and 1 only, we can ignore the first result, sin (x) = -0.25 Using our calculator arcsin (-0.25) = -14.478 Remember that the question wants the value to be between 0 and 360 degrees, we need to find the value that fits within the range Remembering how values of sin repeat themselves, we have 2 answers: 360 - 14.478, or 180 + 14.478 = 345.522, or 194.478