Trigonometry

I knew only that you should make them both the same form. A quick google search of you typed text gave this > http://au.answers.yahoo.com/question/index?qid=20090219015444AA5lj7o

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Walid J answered 5 years ago
OK
sqrt(3)cosx = 1 + sinx

now square both sides u get:

3 cos^2x = sin^2x + 2sinx + 1

but cos^2x = 1 - sin^2x

so 3(1 - sin^2x) = sin^2x + 2sinx + 1

so it becomes:

4sin^2x +2sinx - 2 = 0
div by 2

2sin^2x + sinx - 1 = 0

(2sinx - 1)(sinx + 1) = 0

sinx = 1/2 ---> x = 30 , 150

sinx = -1 ---> x = 3pi/2

Op attributed.

I guess just use most of the answer, and-- aplly the rule for minus not devision etc. If that helps.

sqrt(3)cos(x) - sin(x)= 1

Let's see if this works

(add sin(x) to both sides and square both sides)

3(cos(x)^2 = 1 + 2sin(x) +sin(x)^2

(using cos(x)^2 = 1 - sin(x)^2, we have)

3*(1 - sin(x)^2) = 1 + 2sin(x) +sin(x)^2

(Simplify)

4*sin(x)^2 + 2*sin(x) - 2 = 0

(Divide through by 2)

2*sin(x)^2 = sin(x) -1 = 0

(Factor)

(2sin(x) -1) (sin(x) +1 ) = 0

Solving, we get x = (pi/6 + 2*pi*n , 3*pi/2 +2*pi*n)

Another 2nd quad angle, 5*pi/6, solves (2sin(x) -1 = 0), but it doesn't "work in the original problem) (due to "squaring," we get an extraneous solution??)

I think that may be it.....but check my work and method!!

Thank you Stu and Chris. I really appreciate your answers.

Melody:

sqrt(3)cos(x) - sin(x)= 1

What is the best way to tackle a question like this one? I need a general formula.
Thanks.

(I am reposting because it hasn't been answered yet.)

Here's an alternative approach.

Start by dividing throughout by 2.

(sqrt(3)/2).cosx - (1/2)sinx = 1/2, and then use some well known trig ratios.

sin(pi/3)cosx - cos(pi/3)sinx = 1/2.

Now a trig identity, sin(pi/3 - x) = 1/2,

so pi/3 - x = sin ^-1(1/2) = pi/6 or 5pi/6 (+2k.pi if you wish).

So x = pi/6 or -pi/2 ( + 2k.pi).

Bertie

Your method of substituting cos(pi/3) for 1/2 and sin (pi/3) for (√3)/2 is quite ingenious!!!!

I like that MUCH BETTER !!!

Thanks Bertie, that is what I was hoping for.
I knew there would be something like that I just had a mind blank.

CPhill:

Bertie

Your method of substituting cos(pi/3) for 1/2 and sin (pi/3) for (√3)/2 is quite ingenious!!!!

I like that MUCH BETTER !!!

Sadly, I can't claim to have had any flash of inspiration, the method I used is standard and routine, (with just a tiny realisation).
Post again if you wish/need details.