Question

Tutorial On Finding General Solution for Trigonometric Equations:

Answer 1

General Solution:

Normally, if we are asked to find the solution for \(\theta\) in the interval \((0, 2 \pi)\) for the equation : \(\large \color{red}{\cos(\theta) = \frac{1}{2}}\) then we say \(\large \theta = 60^{\circ} = \frac{\pi}{3}\), but in that interval this is not the only value of \(\theta\), \(\theta\) can have more values like : \(\large \theta = 300^{\circ} = \frac{5 \pi}{3}\)..


As all the trigonometric ratios are cyclic or circular in nature, so we will get infinite values for \(\theta\) that will satisfy the equation \(\large cos(\theta) = \frac{1}{2}\)..


This we can achieve by Finding the General Solution for the given Trigonometric Equation..


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First Form:

1. For \(\color{blue}{sin(\theta) = 0} \) , the general solution for \(\theta\) is : \(\color{blue}{ \theta = n \pi }\) , where \(n \in Z (Integers)\)..

Here, \(n\) can take integer values like -2, -1, 0, 1 , 2 etc..


2. For \(\color{red}{cos(\theta) = 0} \) , the general solution for \(\theta\) is : \(\color{red}{\theta = (2n + 1) \frac{\pi}{2}}\), where \(n \in Z (Integers)\)..

3. \(\color{green}{tan(\theta) = 0 }\) , the general solution for \(\theta\) is : \(\color{green}{\theta = n \pi}\), where \(n \in Z (Integers)\)..

Note: For \(sin(\theta) = 0\) and \(tan(\theta) = 0\) , the general solution is same..


Here is an example:


1. For \(\color{red}{cos(3 \theta) = 0}\) find the general solution..


Solution:

\(cos(3\theta) = 0\)

\(\implies 3 \theta = (2n + 1) \frac{\pi}{2}\) where \(n \in Z\)

\(\implies \theta = (2n + 1) \frac{\pi}{6}\) where \(n \in Z\)



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Second Form:

1. For \(\color{green}{sin(\theta) = sin(\alpha)}\), the general solution is : \(\color{green}{\theta = n \pi + (-1)^{n} \alpha, n

\in Z }\)

2. For \(\color{red}{cos(\theta) = cos(\alpha)}\), the general solution is : \(\color{red}{\theta = 2n \pi \pm \alpha, n \in Z }\)

3. For \(\color{blue}{tan(\theta) = tan(\alpha)}\), the general solution is : \(\color{blue}{\theta = n \pi + \alpha, n \in Z

}\)..



Here is an example:

For \(\color{blue}{cosec(\theta) = 2}\) find the general solution..


Solution:

\(cosec(\theta) = 2\)

\(\implies \large \frac{1}{sin(\theta)} = 2\) or \(sin(\theta) = \frac{1}{2} \implies sin(\theta) = sin(\frac{\pi}{6})\)

So: \(\theta = n \pi + (-1)^{n} (\frac{\pi}{6}) , n \in Z\)


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Third Form:

1. For \(\color{blue}{sin^2(\theta) = sin^2(\alpha)}\) and \(\color{blue}{cos^2(\theta) = cos^2(\alpha)}\) and \(\color{blue}{tan^2(\theta) = tan^2(\alpha)}\), the general solution for the three is same and given by:

\(\color{blue}{\theta = n \pi \pm \alpha}\) where \(n \in Z\)..


Here is an example for this:

For \(\color{orange}{7cos^2(\theta) + 3 sin^2(\theta) = 4}\) find the general solution..


Solution:

\[\color{red}{ 7cos^2(\theta) + 3 sin^2(\theta) = 4}\]

\(\color{blue}{7(1 - sin^2(\theta)) + 3 sin^2(\theta) = 4 \implies 4sin^2(\theta) = 3

\implies sin^2(\theta) = \frac{3}{4}} \)


\(\color{red}{\implies sin^2(\theta) = (\frac{\sqrt{3}}{2})^2 \implies sin^2(\theta) = sin^2 (\frac{\pi}{3})}\)

\(\implies \large \color{green}{\theta = n \pi \pm \frac{\pi}{3}}\) where \(n \in Z\)..

Like wise you can find the general Solution for all the Trigonometric Equations..

Answer 2

i must say good tutorial. and great work.
good for students have difficulties in finding the trigonometric values correctly.

also good latex work:)
you are good in the equation editing tool !

Answer 3

Thanks @sami-21 ..

Answer 4

Some people have curly brown hair,they paint black

some=sine
people have=perpendicular(from people)/hypatenus(from have) so sine=perpendicular/hypatanus

curly=cosine
brown hair=base(from brown)/hypataneus(from hair) so cosine=base/hypatenus

they=tan
paint black=perpendicular(from paint)/base(from black)
so tan=perpendicular/base
whole trignometry in just in line!!!!!!!!!!

Answer 5

@sami-21 check it:))))

Answer 6

i think u also know that @sami-21

Answer 7

\[\huge \text{Nice} \]

Answer 8

\[\huge \color{pink}{Like \quad you...}\]

Answer 9

http://www.interactivetriangulation.com/T/001.html

Answer 10

Let me try something . m quite new to this latex world :P
\[\Huge \color{red} {Awesome!}\]

Answer 11

\[\huge{\mathbb{G}\textbf{r}\mathbb{E}\textbf{a}\mathbb{T}\space \mathbb{W}\textbf{o}\mathbb{R}\textbf{k}}\mathbb{!}\]

Answer 12

good work .. @waterineyes ..

thanks @mathslover for getting me noticed about such a good tutorial

Answer 13

Thanks @vishweshshrimali5 and @mathslover

Answer 14

-_--_--
_____
|-_-|
_|| ||_

:)

Answer 15

\[\huge \color{green}{^{ \cdot}\smile ^{\; \cdot}}\]

Answer 16

Great Work. God bless you :)

Answer 17

@annas thanks buddy..

Answer 18

\[\large{\color{green}{^\cdot}\color{blue}{-}\color{white}{^\cdot}}\]

Answer 19

notice the secnd eye also :)

Answer 20

you're welcome

Answer 21

I am unable to notice @mathslover

Answer 22

That is closed or not ??

Answer 23

select that :)

Answer 24

kush its one eyed smiley :P

Answer 25

:D one green and other one is white ..

notice that :)

Answer 26

Yeah I noticed that..

Ha ha haha..

Answer 27

:D

Answer 28

Just like you..

Thanks @angela210793

Answer 29

And me also..

Answer 30

torture for the epileptics :C lol

Answer 31

yes, very nice indeed.
i especially like "second form" and "third form"

now allow me one small criticism
the line \(\large \theta = 60^{\circ} = \frac{\pi}{3}\)
it is certainly true that the angle whose measure is 60 degrees is the same as the angle whose radian measure is \(\frac{\pi}{3}\) but it is not true that the numbers are the same. as an analogy, the temperature of 77 F is the same as the temperature of 25 C but you would not say \(77=25\)

since as functions of numbers, the trig functions correspond to functions of angles in radians, not degrees, (which is why, for example, the derivative of sine is not cosine, if you are working in degrees) it is best to stick to numbers rather than angles.

Answer 32

I will surely keep it in mind @satellite73

Thanks for your feedback...

Answer 33

awesome

Answer 34

Thanks..!!

Answer 35

gO_Od work!! @eta or @tears