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.

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- Mar 1st 2009, 05:43 AM #1

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## 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.

**PLEASE HELP!!!!!!**

- Mar 1st 2009, 05:58 AM #2

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- Mar 1st 2009, 06:29 AM #3

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- Mar 1st 2009, 06:45 AM #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 and didn't satisfy the original equation, and in the second case and 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 , and then solve .

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

Multiply both sides by :

or .

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

(2) Solve the equation

Square both sides:

This time, this is effectively what sbcd90 did, which is why the extra answers of and were created. In fact these are the solutions to , not (as they should be)

I hope this clears up the mystery.

Grandad

- Mar 1st 2009, 07:26 AM #5

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