cos-1(x) + cos-1(2x) = 60° Let = cos-1(x) Let = cos-1(2x) Then cos() = x and cos() = 2x So the equation cos-1(x) + cos-1(2x) = 60° becomes + = 60° Take the cosine of both sides: cos(+) = cos(60°) Using an identity for cos(+), cos()cos() - sin()sin() = Then since cos() = x and cos() = 2x x(2x) - sin()sin() = 2x² - sin()sin() = Now we have to find the sines. We use the identity: sin² + cos² = 1 sin² = 1 - cos² ___________ sin = ±Ö1 - cos² The sign of the sine depends on what quadrants and are in. We consider all cases ___________ ______ sin(a) = ±Ö1 - cos²(a) = ±Ö1 - x² ___________ __________ _________ sin(b) = ±Ö1 - cos²(b} = ±Ö1 - (2x)² = ±Ö1 - 4x² Substituting in 2x² - sin()sin() = ______ _______ 2x² ± Ö1 - x²·Ö1 - 4x² = _________________ 2x² ± Ö(1 - x²)(1 - 4x²) = Multiply both sides by 2 to clear of the fraction: _________________ 4x² ± 2Ö(1 - x²)(1 - 4x²) = 1 Isolate the radical term: ___________________ ±2Ö(1 - x²)(1 - 4x²) = 1 - 4x² Square both sides: 4(1 - x²)(1 - 4x²) = (1 - 4x²)² 4(1 - 5x² + 4x4) = 1 - 8x² + 16x4 4 - 20x² + 16x4 = 1 - 8x² + 16x4 -12x² = -3 x² = x² = x = ± We must check for extraneous solutions: Checking : cos-1(x) + cos-1(2x) = 60° cos-1() + cos-1(2·) = 60° 60° + cos-1(1) = 60° 60° + 0° = 60° 60° = 60° That checks, so is a solution. Checking : cos-1(x) + cos-1(2x) = 60° cos-1() + cos-1(2·) = 60° 120° + cos-1(-1) = 60° 120° + 180° = 60° 300° = 60° That's false, so the only solution is x = Edwin
