When an equation of the form a sin θ + b cos θ is equated to 0, is it possible to divide throughout by cos θ, so that the original trigonometric equation is reduced to a trigonometric equation in terms of tan θ? Or would you be losing a solution in doing so?
Can anyone tell me what's wrong with the solution of the equation below? I know that by dividing throughout by cos θ, I would have got to the answer, but what's wrong with the method below?
Solve 3^0.5 sin θ - cos θ = 0 for 0º<θ<360º
Since sin θ = cos θ tan θ, I worked it out as follows:
3^0.5 cos θ tan θ - cos θ = 0
cos θ (3^0.5 tan θ - 1) = 0
so either cos θ = 0 or 3^0.5 tan θ - 1 = 0
θ = 90º or θ = 30º
Then by writing the general solutions to both, I found out that the angles θ within the range are θ = 0º, 30º, 210º, 270º
First problem: the book which I am using only gives out θ = 30º and 210º as answers and teacher said there are no mistakes with answers
Second problem: When I substituted the values of θ =0º and 270º in the original equation, I found out that these are not correct values since 3^0.5 sin 0 - cos 0 is not equal to 0.