Trigonometric equations

Trigonometric equations

Equation involving trigonometric functions defined above are call trigonometric equation. Since we know that value of sin and cos function repeat after 2π interval and value of tan function repeat after π interval.So, there will be infinite solution for these equations.
We define two forms of solutions here
Principle Solution: The solution in the range 0 ≤ x ≤ 2π

Check out below links on how to remember the trigonometry tables
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General Solution: The expression involving integer n which gives all solutions of a trigonometric equation is called the general solution.
We shall use 'Z' to denote the set of integers.

We already know the general solutions of the below trigonometric equations
$\sin (x) = 0$ implies $x = n \pi/2$, where n is any integer
$\cos (x) = 0$ implies $x = (2n + 1) \frac {\pi}{2}$

Theorem -1
sinx =siny then $x=n \pi + (-1)^{n}y$ where n is any integer
Proof
If sin x = sin y, then
$\sin x - \sin y = 0$ or
$2 \cos \frac {x+y}{2} \sin \frac {x-y}{2} =0$
Which means
$\cos \frac {x+y}{2}=0$ or $\sin \frac {x-y}{2}=0$
for $\cos \frac {x+y}{2}=0$
$ \frac {x+y}{2} = (2n+1) \frac {\pi}{2} $ where $n \in Z$
i.e. $x = (2n + 1) \pi - y $
for $\sin \frac {x-y}{2}=0$
or $x = 2n \pi + y$ where $n \in Z$
Hence $ x = (2n + 1)\pi + (-1)^{2n + 1} y$ or $x = 2n \pi +(-1)^{2n} y$ where $n \in Z$
Combining these two results, we get
$x = n \pi + (-1)^{n} y$, where $n \in Z$.

Theorem 2
cosx=cosy then $x=2n \pi + y$ or $x=2n \pi - y$ where n is any integer
Proof
if cosx=cosy
cosx - cos y=0
$ -2 \sin \frac {x+y}{2} \sin \frac {x-y}{2}=0$
Which means
$\sin \frac {x+y}{2} =0$ or $\sin \frac {x-y}{2}=0$
for $\sin \frac {x+y}{2} =0$
$ \frac {x+y}{2}= n \pi $
or $x = 2n \pi -y$
For $\sin \frac {x-y}{2}=0$
$ \frac {x-y}{2}= n \pi $
or $x = 2n \pi +y$
Combining these two results
$x=2n \pi + y$ or $x=2n \pi - y$ where n is any integer

Theorem 3
tanx=tany then $x=n \pi +y$ or $x=n \pi- y$ where n is any integer
Proof
tan x -tan y=0
$\frac {\sin(x) \cos(y) - \sin(y) \cos(x)}{\cos(x) \cos(y)}=0$
$ \frac {\sin (x-y)}{\cos(x) \cos (y)}=0$
Which means sin (x-y) =0
$x- y = n \pi$
or $ x= n\pi + y$

How to Find general solution of trigonometric equations

1. Obtain the equation in the form given above
i.e
sin x ="value"
cos x="value>"
tan x="value>"

2.Convert into the below forms
sinx =sin y
cos x=cos y
tan x=tan y
We can use first principle value of x as y. It is not mandatory , we can choose any y which satisfies the value

3. Use the below formula to find the general solutions of the trigonometric equations
sinx =siny then $x=n \pi + (-1)^{n}y$ where n is any integer
cosx=cosy then $x=2n \pi + y$ or $x=2n \pi - y$ where n is any integer
tanx=tany then $x=n \pi +y$ where n is any integer

Some basics Tips to solve the trigonometric equations questions

(1) Always try to bring the multiple angles to single angles using basic formula.Make sure all your angles are the same. Using sin(2x) and sinx is difficult, but if you use $\sin2x = 2 \sin(x) \cos(x)$, that leaves sin(x) and cos(x), and now all your functions match.
The same goes for addition and subtraction: don't try working with sin(x+y) and sinx. Instead, use $\sin(x+y) = \sin(x) \cos(y)+\cos(x) \sin(y)$ so that all the angles match
(2) Converting to sin and cos all the items in the problem using basic formula. I have mentioned sin and cos as they are easy to solve.You can use any other also.
(3) Check all the angles for sums and differences and use the appropriate identities to remove them.
(4) Use Pythagoras identifies to simplify the equations
(5) For Equation containing higher degree of sinx or cos x, try to convert into linear form using the below formula
$\sin^2(x) = \frac {1- \cos(2x)}{2}$
$\cos^2(x) = \frac { 1+ \cos(2x)}{2}$
$\sin^3(x)= \frac {3 \sin(x) -\sin(3x)}{4}$
$\cos^3(x)= \frac { 3\cos (x) + \cos (3x)}{4}$
(6) Practice and Practice. You will soon start figuring out the equation and there symmetry to resolve them fast

Mistakes which should be avoided
a. Squaring the trigonometric equation should be avoided as it gives extraneous values
b.Never cancel terms containing the variable on both sides as you may loose the genuine solutions
c.Often the equation may give two solution sets which may be disjoint, overlapping, subsets
i. For disjoint, provide both the solutions
ii.For overlapping one , try to write the solutions set in the form such that common solution is in one set only
iii.For subsets, we can provide the final super set in the solution only

d.If tan x or sec x is involved in the equation, $x \neq (2n+1) \frac {\pi}{2}$ as it is undefined.So make sure to check the solutions for these values in case tan or sec function are involved
e.If cotx or cosec x is involved in the equation, $x \neq n \pi \; or \; 0$ as it is undefined.So make sure to check the solutions for these values in case cotx or cosec function are involved

Important forms of Trigonometric Equations

(a) Equation of the form
$a \cos(x) + b \sin(x) =c$ such that $|c| \leq \sqrt {a^2 + b^2}$

We can solve it using the below method
i. Put $a =r \cos (\alpha)$ and $b=r \sin (\alpha)$ where $r=\sqrt {a^2 + b^2}$ and $ \tan (\alpha)= \frac {b}{a}$
ii. This reduces the above equation to the form
$r \cos(x) \cos( \alpha) + r \sin (x) \sin (\alpha)=c$
$r \cos (x - \alpha)=c$
$ \cos (x - \alpha ) = \frac {c}{r}$
Now this can be easily solved using the general solution given above

(b) Equation of the form $\sin^2 (x) = \sin^2 (y), \cos^2 (x) = \cos^2 (y) , \tan^2 (x) = \tan^2 (y)$ General solution is given by $x = n \pi \pm y$ where n is any integer
(c) Equation of the form
$a \sin^2 (x) + b \sin (x) cos(x) + c \cos^2 (x) + d=0$

We can solve using the below method
i. if $a + d = 0$,then cos x =0 is solution of the equation,
$-d \sin^2 (x) + b \sin (x) \cos (x) + c \cos^2 (x) + d=0$
$(c+d) \cos^2 (x) + b \sin (x) \cos (x)$
Which can be solved easily

ii. if $a + d \neq 0$,then cos x =0 is not solution of the equation
We can divide the equation by $\cos^2(x)$
$a \tan^2 (x) + b \tan (x) + c + d \sec^2 (x)=0$
Now $\sec^2 (x)= 1+ \tan^2 (x)$
So,
$a \tan^2 (x) + b \tan (x) + c + d(1 + \tan^2 (x))=0$
$(a+d) \tan^2 (x) + b \tan(x) +c+d=0$
which is a quadratic equation in tan x and solutions can be obtained using easily

(d) Equation of the form
|sin x|=1 ,General solution is given by $x= (2n+1) \frac {\pi}{2}$
|cos x|=1,General solution is given by $x=n \pi$




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