# Thread: a sin θ + b cos θ = 0

1. ## a sin θ + b cos θ = 0

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.

2. 3^0.5 sin A=cosA

or,3sin^2A=cos^2A

or,4 sin^2 A=1

or,sin A=(+-)1/2

so,A=30,150,210,330deg.

but,A=150,330 deg. are trivial cases.

hence,A=30,210.

3. Many thanks for the alternative correct methods, but I have two queries:

what's the problems with the method I used?
How have you decided that 150º is a trivial angle?

Thanks a lot

4. ## Trig equations

Hello yobacul and sbcd90

You have to be a little more careful when you're solving equations that you don't introduce solutions that aren't really there. Both of you have done that in what you have written. In the first case the values $\theta = 0^o$ and $270^o$ didn't satisfy the original equation, and in the second case $A = 150^o$ and $330^o$ didn't either.

So, there are two questions to be answered:

• What is the correct way of solving this equation?
• Where have these extra answers come from?

The answer to the first question is the one that yobacul gave at the beginning: divide throughout by $\cos\theta$, and then solve $\sqrt 3\tan\theta - 1 = 0$.

To answer the second question (and there are really two questions here, because both alternative methods introduced extra, unwanted, answers) consider the following examples:

(1) Solve the equation $2x - 3 = 0$

Multiply both sides by $x$:

$x(2x - 3) = 0$

$\Rightarrow x = 0$ or $1.5$.

That's rather silly, isn't it? But that's effectively what yobacul did when you multiplied by $\cos\theta$, and then said that $\cos \theta = 0$ is a possible solution!

(2) Solve the equation $5x = 10$

Square both sides:

$25x^2 = 100$

$\Rightarrow x^2 = 4$

$\Rightarrow x = \pm 2$

This time, this is effectively what sbcd90 did, which is why the extra answers of $150^o$ and $330^o$ were created. In fact these are the solutions to $\tan\theta = -\frac{1}{\sqrt 3}$, not (as they should be) $\tan\theta = \frac{1}{\sqrt 3}$

I hope this clears up the mystery.