1. 2 sin x + tan x =0

2. sin x =1

3. sinx sin 3x =1/2

4. sin x=0

5. sin x + cos x =0

We define two forms of solutions here2. sin x =1

3. sinx sin 3x =1/2

4. sin x=0

5. sin x + cos x =0

1. Find the Principal solutions of the below trigonometric equations

a. sin x= 1/2

b. cos x =-1/2

c. tan x=-1

**Solutions**
The principle solutions lies in the range 0 ≤ x ≤ 2π. We can principle solution using the below method

i. Find the value of the angle using the postive value.

ii) Then remember these below identitites to find the principle values

$sin( \pi -x )=sin x$

$sin ( \pi + x) = -sinx $

$sin ( 2\pi - x) = -sinx $

$cos ( \pi -x )= -cos x$

$cos ( \pi +x )= -cos x$

$cos ( 2\pi -x )= cos x$

$tan ( \pi -x )= -tan x$

$tan ( \pi +x )= tan x$

$tan ( 2\pi -x )= -tan x$

a) sin x=1/2

$sin x= sin (\pi/6)$

Now $sin (\pi - \pi/6) = sin \pi/6$

or $ sin (5 \pi/6) = 1/2$

So principle solutions are π/6, 5π/6

b) cos x =-1/2

$cos x = -cos (\pi /3)$

Now $cos (\pi - \pi /3) = -cos (\pi /3)$

or $cos (2 \pi/3) = -1/2$

Also $cos (\pi + \pi /3)= -cos (\pi /3)$

or $cos (4 \pi/3) = -1/2$

So principle solutions are 2π/3, 4π/3

c) tanx =-1

$ tan x = -tan (\pi/4)$

Now $ tan ( \pi -\pi/4 )= -tan (\pi/4)$

or $ tan 3\pi/4 = -1$

Also $tan ( 2\pi - \pi/4 )= -tan \pi/4$

or $ tan 7 \pi/4 = -1$

So principle solutions are 3π/4, 7π/4

a. sin x= 1/2

b. cos x =-1/2

c. tan x=-1

i. Find the value of the angle using the postive value.

ii) Then remember these below identitites to find the principle values

$sin( \pi -x )=sin x$

$sin ( \pi + x) = -sinx $

$sin ( 2\pi - x) = -sinx $

$cos ( \pi -x )= -cos x$

$cos ( \pi +x )= -cos x$

$cos ( 2\pi -x )= cos x$

$tan ( \pi -x )= -tan x$

$tan ( \pi +x )= tan x$

$tan ( 2\pi -x )= -tan x$

a) sin x=1/2

$sin x= sin (\pi/6)$

Now $sin (\pi - \pi/6) = sin \pi/6$

or $ sin (5 \pi/6) = 1/2$

So principle solutions are π/6, 5π/6

b) cos x =-1/2

$cos x = -cos (\pi /3)$

Now $cos (\pi - \pi /3) = -cos (\pi /3)$

or $cos (2 \pi/3) = -1/2$

Also $cos (\pi + \pi /3)= -cos (\pi /3)$

or $cos (4 \pi/3) = -1/2$

So principle solutions are 2π/3, 4π/3

c) tanx =-1

$ tan x = -tan (\pi/4)$

Now $ tan ( \pi -\pi/4 )= -tan (\pi/4)$

or $ tan 3\pi/4 = -1$

Also $tan ( 2\pi - \pi/4 )= -tan \pi/4$

or $ tan 7 \pi/4 = -1$

So principle solutions are 3π/4, 7π/4

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 $x = (2n + 1) \frac {\pi}{2}$

sinx =siny then $x=n \pi + (-1)^{n}y$ where n is any integer

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 $n \in Z$.

cosx=cosy then $x=2n \pi + y$ or $x=2n \pi - y$ where n is any integer

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

tanx=tany then $x=n \pi +y$ or $x=n \pi- y$ where n is any integer

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$

i.e

sinx =

cos x=

tan x=

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

a. $sin x = - \frac {1}{2}$

b. $cos x = \frac {\sqrt {3}}{2}$

c. $cos x = \frac {1}{2}$

we will use the same steps as described above

a. $sin x = - \frac {1}{2}$

$sin x = sin (\pi + \pi/6) = sin (7 \pi/6)$

So general solution would be

$x=n \pi + (-1)^{n} \frac {7 \pi}{6}$ where n is any integer

b. $cos x = \frac {\sqrt {3}}{2}$

$cos x = cos (\pi/6)$

So general solution would be

$x=2n \pi + \frac {\pi}{6}$ or $x=2n \pi - \frac {\pi}{6}$ where n is any integer

c.$cos x = \frac {1}{2}$

$cos x = cos (\pi/3)$

So general solution would be

$x=2n \pi + \frac {\pi}{3}$ or $x=2n \pi - \frac {\pi}{3}$ where n is any integer

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 phythagorean identifies to simplfy the equations

4) For Equation containing higher degree of sinx or cos x, try to convert into linear form using the below formula

$sin^2 x = \frac {1- cos2x}{2}$

$cos^2x = \frac { 1+ cos2x}{2}$

$sin^3 x= \frac {3sinx -sin 3x}{4}$

$cos^3 x= \frac { 3cos x + cos 3x}{4}$

5) Practice and Practice. You will soon start figuring out the equation and there symmetry to resolve them fast

b.Never cancel terms containing the variable on both sides as you may loose the geninue 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

$2 cos^2 (x)+ 3 sin(x)=0$

$2 cos^2(x)+ 3 sin(x)=0$

Now first convert inton one function using trigonometric identities

$sin^2 x + cos^2 x=1$

So,

$2(1 -sin^2 x) + 3 sinx =0$

$2 sin^2 x-3 sinx-2=0$

Solving the above quadratic equation,

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

or sin x = -1/2 as $sinx \neq 2$

Now

sin x = -1/2

$sin x = sin (\pi + \pi/6) = sin (7 \pi/6)$

So general solution would be

$x=n \pi + (-1)^{n} \frac {7 \pi}{6}$ where n is any integer

$tan 2x = -cot (x + \pi/3)$

$tan 2x = -cot (x + \pi/3)$

$tan 2x = tan ( \frac {\pi}{2} +x + \frac {\pi}{3})$

$tan 2x = tan ( \frac {5\pi}{6} +x )$

So,

$2x = n \pi + \frac {5\pi}{6} +x$

$x= n \pi + \frac {5\pi}{6}$

$ tan x + tan 2x - tan (x) tan(2x) tan(3x) + tan 3x=0$

$ tan x + tan 2x =-tab 3x( 1 -tan (x) tan (2x))$

$ \frac {tan x + tan 2x}{1-tan (x) tan (2x)}= -tan 3x$

$ tan 3x =-tan 3x$

or tan 3x=0

General solution,

$3x =n \pi$

$x = \frac {n \pi}{3}$ for $n \in Z$

a. Equation of the form

$a cos(x) + bsin(x) =c$ such that $|c| \leq \sqrt {a^2 + b^2}$

We can solve it using the below method

i. Put $a =r cos \alpha$ and $b=rsin \alpha$ 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 $sin^x = sin^2 y, cos^2 x = cos^2 y , tan^2 x = tan^2 y$ General solution is given by $x = n \pi \pm y$ where n is any integer

c. Equation of the form

$a sin^2 x + bsin (x) cos(x) + c cos^x + d=0$

We can solve using the below method

i. if $a + d = 0$,then cos x =0 is solution of the equation,

$-d sin^2 x + bsin x cos x + c cos^x + d=0$

$(c+d) cos^x + bsin x cos x$

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 $cos^2x$

$a tan^2 x + b tan x + c +dsec^2 x=0$

Now $sec^2 x= 1+ tan^2x$

So,

$a tan^2 x + b tan x + c + d(1 + tan^2 x)=0$

$(a+d)tan^2x + b tanx +c+d=0$

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 $x=n \pi$

Solve $cot x + cosec x = \sqrt {3}$

$ \frac {cos x}{sin x} + \frac {1}{sin x} =\sqrt {3}$

or

$\sqrt {3} sin x - cos x =1$

This is of the form $a cosx + bsin x =c$

We have a =-1 ,$b= \sqrt {3}$, c=1

$\sqrt {3} = r sin \alpha$ and $ 1 = r cos \alpha$

So $tan \alpha = \sqrt {3} = tan ( \pi/3)$ and $r = \sqrt { 3 + 1} =2$

So finally we get

$2 sin (\pi /3) sin x - 2 cos (\pi/3) cos x =1$

$ -2 cos (x + \pi /3) =1$

$ cos (x + \pi /3)= -1/2$

$ cos (x + \pi /3)= cos (2 \pi /3)$

General solutions

$ x + \frac {\pi}{3} = 2n \pi \pm \frac {2 \pi}{3}$ for $n \in Z$

$ x= 2n \pi \pi \frac {2 \pi}{3} - \frac {\pi}{3}$

Taking Positive sign

$ x= 2n \pi + \frac {\pi}{3}$

Taking negative sign

$x = 2n \pi - \pi = (2n-1) \pi $

But this is undefined as cot x and cosec x are defined for multiple of $\pi$

So solution set is

$ x= 2n \pi + \frac {\pi}{3}$ for $n \in Z$

1. $ \sqrt {2} sec x =2$

2. $sin 2x - sin 4x + sin 6x = 0$

3. $ \sqrt {3} cos x + sin x = \sqrt {2}$

4. $ sin^2 x = \frac {3}{4}$

5. $cos \theta + cos 3 \theta + cos 5 \theta + cos 7 \theta =0$

Answers

Class 11 Maths Class 11 Physics