6 cos(θ) + 9 sin(θ) = 2

Question

solve algebraically

Answer 1

Multiply and divide the left side by √(6²+9²) = √117 to get:

 

√117[(6/√117)cosθ + (9/√117)sinθ] = 2

 

          Draw a right triangle with one acute angle labelled α, side adjacent to α of length 9 side opposite angle α of length 6, and            hypotenuse of length √117.

 

We then have:  sinαcosθ + cosαsinθ = 2/√117

 

                So, sin(α+θ) = 2/√117

 

                      Since α is acute and sinα = 6/√117, we have α = Arcsin(6/√117) = 33.69°

 

                      sin(33.69° + θ) = 2/√117

 

                            θ + 33.69° = Arcsin(2/√117) + k(360°) or (180° - Arcsin(2/√117)) + k(360°)

 

                                             = 10.66° + k(360°) or 169.34° + k(360°)

 

                                        θ = -23.03° + k(360°) or 135.65° +k(360°)  k = 0, ±1, ±2, ...

 

For example, when k = 0, we get θ = -23.03° or 135.65°

 

 

 

 

Answer 2

begin by replacing cosθ=±√(1-sin²θ). then isolate the radical, then square both sides, and you will have an equation that will be manageable.  you will definitely have to check your answers!