Solve 9 sin2 Ө – 9 sin Ө = 11

Last night at Mont Kiara, the above question was forwarded to me to solve.

Anyway, you will note from my proposed solutions below that different domains for Ө will mean different answers. 

For examples:

* For 0 ≤ Ө ≤ π radians, no value of Ө can satisfy the equation: 9 sin² Ө – 9 sin Ө = 11;

* For 0 ≤ Ө ≤ 2π radians, two values Ө satisfy the equation

My Proposed Solutions

                                  9 sin² Ө – 9 sin Ө = 11

-11:                   9 sin² Ө – 9 sin Ө – 11 = 0

                        (compare: ax² + bx + c = 0)

where:                   a = 9, b = -9 and c = -11

therefore, roots:     [α = (-b + √(b² – 4ac))/2a; or β = (-b - √(b² – 4ac))/2a]

imply:          sin Ө = -(-9) - √(-9)² – 4(9)(-11))/(2x9) = - 0.713351648; or,

                    sin Ө = -(-9) + √(-9)² – 4(9)(-11))/(2x9) =  1.713351648;

For 0 ≤ Ө ≤ 2π radians:

                    sin Ө = - 0.713351648 implies Ө in 3^rd or 4^th quadrants

                          Ө = π + sin^-1 (0.713351648) or 2π - sin^-1 (0.713351648)

                          Ө = 3.935861836 (3rd Qd) or 5.488916105 (4th Qd)

                          Ө = 3.94 or 5.49 rad (both to 3 sf)

But for 0 ≤ Ө ≤ π radians (1^st and 2^nd quadrants):

                    For all values of Ө, sin Ө must be +ve.

                    Therefore, sin Ө ≠ - 0.713351648.

Also for all values of Ө in any quadrant, sin Ө ≠ 1.713351648 because 1 ≥  sin Ө ≥ -1.

So remember: If you want to solve a trigonometric equation, you must know or be given the domain because the answers depend on the domain for Ө! In fact, Q 10(e) of Edexcel Further Pure Maths (4PMO) Yr 2013 May Paper 1 amply demonstrates this!!