Mastering "The 't' Method": Solving Trig Equations w/ Pavadrin

[pavadrin]

“The “t” Method”

Hey
Recently I have been studying for an upcoming test where it requires me to use “The “t” Method”. In this method the value of x for trigonometric equations is determined through vair the key component of “The “t” Method” is:
t=tan \frac{A}{2}
tan A=\frac{1+t^2}{1-t^2}
sin A=\frac{2t}{1+t^2}
cos A=\frac{1-t^2}{1+t^2}
If this is famliar to a reader by another name, could you please post this methods other name.

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Problems which i need to solve:
1. 2sin x + 3cos x = 5
2. 3tan x + \sqrt{3}sec x=1
3. 10cos (pi x) + 3sin (2pi x)=4
4. 3sin 2x + 5cot 3x = 7
5. csc x + 2sec (pi x)

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My working for question 1. 2sin x + 3cos x = 5
2\frac{2t}{1+t^2} + 3\frac{1-t^2}{1+t^2} = 5
\frac{4t + 3 - 3t^2}{1+t^2} = 5
4t + 3 -3t^2 = 5 +5t^2
4t + 2t^2 = 2
2t^2 + 4t - 2 = 0
t = 0.4142135624 or t = -2.414213562
t = \tan\frac{x}{2}
x = 2tan^-1 0.4142135624 or x = 2tan^-1 -2.414213562
x = 45 or x = -135.0005034

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I conclude my working here as I am not sure if it is correct. Is there an aleternaitve method I could use to check these final answers? And are there any more values of x for which the equation is true?
Thanks well in advance,
Pavadrin

 

[HallsofIvy]

Hey
Recently I have been studying for an upcoming test where it requires me to use “The “t” Method”. In this method the value of x for trigonometric equations is determined through vair the key component of “The “t” Method” is:
t=tan \frac{A}{2}
tan A=\frac{1+t^2}{1-t^2}
sin A=\frac{2t}{1+t^2}
cos A=\frac{1-t^2}{1+t^2}
If this is famliar to a reader by another name, could you please post this methods other name.

__________________________________________________​

Problems which i need to solve:
1. 2sin x + 3cos x = 5
2. 3tan x + \sqrt{3}sec x=1
3. 10cos (pi x) + 3sin (2pi x)=4
4. 3sin 2x + 5cot 3x = 7
5. csc x + 2sec (pi x)

__________________________________________________​

My working for question 1. 2sin x + 3cos x = 5
2\frac{2t}{1+t^2} + 3\frac{1-t^2}{1+t^2} = 5
\frac{4t + 3 - 3t^2}{1+t^2} = 5
4t + 3 -3t^2 = 5 +5t^2
4t + 2t^2 = 2

You've lost track of a sign: adding 3t² to both sides gives 8t^2- 4t+ 2= 0 or 4t^2- 2t+ 1= 0
That has only complex roots.

In fact, since 2+ 3= 5 and sine and cosine are never larger than 1, the only way we could have 2sin x+ 3cos x= 5 is to have sin x= 1 and cos x=1 which is not true for any x.

 

[pavadrin]

thanks HallofIvy.
do they all have complex roots, or do i need to wokr them as i have done before to determine x when i have it in terms of 2tan^-1 t ?

 

[HallsofIvy]

Well, I don't know! I only looked at the first as that was the one for which you showed your work. I imagine that at least some of these have real solutions but you will have to do the algebra to see.

 

[pavadrin]

oh okay, ill post my wokring for the otherss when i have more time.

 

[pavadrin]

working for question number 2.

3tan x + \sqrt{3}sec x=1

3\frac{2t}{1-t^2} + \sqrt{3}\frac{1+t^2}{1-t^2} = 1
\frac{6t + \sqrt{3} + t^2\sqrt{3}}{1 - t^2} = 1
6t + \sqrt{3} + t^2\sqrt{3} = 1 - t^2
(1 + \sqrt{3})t^2 + 6t - (1 - \sqrt{3}) = 0
therefore: t = -0.1296640194 or t = -2.066488403
for t = -0.1296640194
t = tan^-1 \frac{x}{2}
x = 2tan^-1 -0.1296640194
x = -14.77596194
t = -2.066488403
t = tan^-1 \frac{x}{2}
x = 2tan^-1 -2.066488403
x = -128.3541404




again, I am not sure if what i am do is right so i would appreciate it iof somebody helped me out. if anybody knows of alternative ways of solving these equations, please tell me how

many thanks
Pavadrin.

 

[0rthodontist]

This is also known as the Weierstrass substitution.

 

[eglipo]

Hey
Recently I have been studying for an upcoming test where it requires me to use “The “t” Method”. In this method the value of x for trigonometric equations is determined through vair the key component of “The “t” Method” is:
t=tan \frac{A}{2}
tan A=\frac{1+t^2}{1-t^2}
sin A=\frac{2t}{1+t^2}
cos A=\frac{1-t^2}{1+t^2}
If this is famliar to a reader by another name, could you please post this methods other name.

__________________________________________________​

Problems which i need to solve:
1. 2sin x + 3cos x = 5
2. 3tan x + \sqrt{3}sec x=1
3. 10cos (pi x) + 3sin (2pi x)=4
4. 3sin 2x + 5cot 3x = 7
5. csc x + 2sec (pi x)

__________________________________________________​

My working for question 1. 2sin x + 3cos x = 5
2\frac{2t}{1+t^2} + 3\frac{1-t^2}{1+t^2} = 5
\frac{4t + 3 - 3t^2}{1+t^2} = 5
4t + 3 -3t^2 = 5 +5t^2
4t + 2t^2 = 2
2t^2 + 4t - 2 = 0
t = 0.4142135624 or t = -2.414213562
t = \tan\frac{x}{2}
x = 2tan^-1 0.4142135624 or x = 2tan^-1 -2.414213562
x = 45 or x = -135.0005034

__________________________________________________​

I conclude my working here as I am not sure if it is correct. Is there an aleternaitve method I could use to check these final answers? And are there any more values of x for which the equation is true?
Thanks well in advance,
Pavadrin

You’ve made a mistake between the third and forth lines. The moves to the left with a minus, so as a result you get , which has no real solutions use quadratic formula).

Alternative approach to trig

 

[pavadrin]

okay,many thanks once again for the replies