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Trigonometric equations involve trigonometric functions that are periodic and therefore their solutions can occur in one or two quadrants and all revolutions. On every quadrant, trigonometric functions behave differently. In some instances, they might behave the same but there is no hard and fast rule to understand this concept. To understand trigonometric equations, you need to understand how trigonometric functions behave in different quadrants. Furthermore, you will also require to understand fundamental trigonometric identities. These identities will help you to make your equation easier to understand and solve. Does this mean that you need to understand all the trigonometric identities? Yes! You never know which identity is required hence having knowledge of all fundamental trigonometric identities is important to solve a trigonometric equation. In addition, once you solved the trigonometric equation, you will be left with an angle or sometimes you might be left with more than one angle. That is never the endpoint, with the help of that angle, we find more angles by drawing 4 quadrants and then marking all the possibilities of that angle to lie.

Usually, we try to find the angle within the { 360 }^{ \circ } or 2 \pi (if you are working in radians). The reason is that the repetition starts after { 360 }^{ \circ } or 2 \pi hence prediction becomes easier that is why we usually work till { 360 }^{ \circ } or 2 \pi but that doesn't mean we don't proceed from { 360 }^{ \circ } or 2 \pi, sometimes we do and sometimes we move in the reverse direction as well (meaning all the angle will be negative). Below are some examples, if you have a good understanding of trigonometric identities and the behavior of trigonometric functions in different quadrant then why don't you give them a try? And in case if you got stuck in some, we have already provided the solution and then you can check where you made the mistake or you can tally the answer as well.

Examples

Solve the trigonometric equation:

1.2 \tan { x } - 3 \cot { x } - 1 = 0

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TRIGONOMETRIC IDENTITIES USED:

\cot { x } = \frac { 1 }{ \tan { x } }

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2 \tan { x } - \frac { 3 }{ \tan { x } } - 1 = 0

2 \tan^{ 2 }{ x } - \tan { x } - 3 = 0

Assuming that \tan { x } = y:

2 { y }^{ 2 } - y - 3 = 0

y = \frac { -(-1) \pm \sqrt { { (-1) }^{ 2 } - 4 (2)(-3) } }{ 2(2) }

y = \frac { 1 \pm \sqrt {  1 + 24 } }{ 4 }

y = \frac { 1 \pm 5 }{ 4 }

y = \frac { 1 + 5 }{ 4 } \qquad y = \frac { 1 - 5 }{ 4 }

y = \frac { 6 }{ 4 } \qquad y = \frac { -4 }{ 4 }

y = \frac { 3 }{ 2 } \qquad y = -1

\tan { x } = \frac { 3 }{ 2 } \qquad \tan { x } = -1

x = { 56.30 }^{ \circ } \qquad x = { 135 }^{ \circ }

For x = { 56.30 }^{ \circ }:

x = { 56.30 }^{ \circ } + { 180 }^{ \circ }k

 

For x = { 135 }^{ \circ }:

x =  { 135 }^{ \circ } + { 180 }^{ \circ }k

 

 

2.\cos^{ 2 }{ x } - 3 \sin^{ 2 }{ x } = 0

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TRIGONOMETRIC IDENTITIES USED:

\sin^{ 2 }{ x } + \cos^{ 2 }{ x } = 1 \Rightarrow \cos^{ 2 }{ x } = 1 - \sin^{ 2 }{ x }

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1 - \sin^{ 2 }{ x } - 3 \sin^{ 2 }{ x } = 0

1 - 4 \sin^{ 2 }{ x } = 0

{ (1) }^{ 2 } - { (2 \sin { x }) }^{ 2 } = 0

(1 - 2 \sin { x })(1 + 2 \sin { x }) = 0

1 - 2 \sin { x } = 0 \qquad 1 + 2 \sin { x } = 0

\sin { x } = \frac { 1 }{ 2 } \qquad \sin { x } = - \frac { 1 }{ 2 }

x = \arcsin { \frac { 1 }{ 2 } } \Rightarrow \left\{\begin{matrix} { x }_{ 1 } = { 30 }^{ \circ } + { 360 }^{ \circ }k \\ { x }_{ 2 } = { 150 }^{ \circ } + { 360 }^{ \circ }k \end{matrix}\right

x = \arcsin { - \frac { 1 }{ 2 } } \Rightarrow \left\{\begin{matrix} { x }_{ 1 } = { 210 }^{ \circ } + { 360 }^{ \circ }k \\ { x }_{ 2 } = { 330 }^{ \circ } + { 360 }^{ \circ }k \end{matrix}\right

 

 

3.\sin { (2x + { 60 }^{ \circ }) + \sin { (x + { 30 }^{ \circ }) }} = 0

Sum to Product:

2 \sin { (\frac { 3x }{ 2 } + { 45 }^{ \circ }) } \cos { (\frac { x }{ 2 } + { 15 }^{ \circ }) } = 0

Divide by 2 and equate each factor to 0.

\sin { (\frac { 3x }{ 2 } + { 45 }^{ \circ }) } = 0 \Rightarrow \left\{\begin{matrix} \frac { 3x }{ 2 } + { 45 }^{ \circ } = { 0 }^{ \circ } + { 360 }^{ \circ }k \qquad x = -{ 30 }^{ \circ } + { 120 }^{ \circ }K \\ \frac { 3x }{ 2 } + { 45 }^{ \circ } = { 180 }^{ \circ } + { 360 }^{ \circ }k \qquad x = - { 30 }^{ \circ } + { 120 }^{ \circ }K \end{matrix}\right

\cos { (\frac { x }{ 2 } + { 15 }^{ \circ }) } = 0 \Rightarrow \left\{\begin{matrix} \frac { x }{ 2 } + { 15 }^{ \circ } = { 90 }^{ \circ } + { 360 }^{ \circ }k \qquad x = { 150 }^{ \circ } + { 360 }^{ \circ }K \\ \frac { x }{ 2 } + { 15 }^{ \circ } = { 270 }^{ \circ } + { 360 }^{ \circ }k \qquad x =  { 510 }^{ \circ } + { 360 }^{ \circ }K \rightarrow x = { 150 }^{ \circ } + { 360 }^{ \circ }K \end{matrix}\right

 

4.\sin^{ 2 }{ x } - \cos^{ 2 }{ x } = \frac { 1 }{ 2 }

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TRIGONOMETRIC IDENTITIES USED:

\cos { 2x } = \cos^{ 2 }{ x } + \sin^{ 2 }{ x }

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\sin^{ 2 }{ x } - \cos^{ 2 }{ x } = \frac { 1 }{ 2 }

-(\cos^{ 2 }{ x } + \sin^{ 2 }{ x }) = \frac { 1 }{ 2 }

\cos^{ 2 }{ x } + \sin^{ 2 }{ x } = - \frac { 1 }{ 2 }

\cos { 2x } = - \frac { 1 }{ 2 }

2x = \left\{\begin{matrix} { 120 }^{ \circ } + { 360 }^{ \circ }k \\ { 240 }^{ \circ } + { 360 }^{ \circ }k \end{matrix}\right

x = \left\{\begin{matrix} { 60 }^{ \circ } + { 180 }^{ \circ }k \\ { 120 }^{ \circ } + { 180 }^{ \circ }k \end{matrix}\right

 

 

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

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TRIGONOMETRIC IDENTITIES USED:

\sin { \alpha + \beta } = \sin { \alpha } \cos { \beta } + \cos { \alpha } \sin { \alpha }

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

\frac { 1 }{ 2 } \sin { x } + \frac { \sqrt { 3 } }{ 2 } \cos { x } = 1

\cos { { 60 }^{ \circ } } \sin { x } + \sin { { 60 }^{ \circ } } \cos { x } = 1

\sin { x + { 60 }^{ \circ } } = 1

x + { 60 }^{ \circ } = { 90 }^{ \circ } + { 360 }^{ \circ } k

x = { 30 }^{ \circ } + { 360 }^{ \circ } k

6.\sin { 2x } = \cos { { 60 }^{ \circ } }

\sin { 2x } = \frac { 1 }{ 2 }

\left\{\begin{matrix} 2x  = { 30 }^{ \circ } + { 360 }^{ \circ }k \qquad x = { 15 }^{ \circ } + { 180 }^{ \circ } k \\ 2x = { 150 }^{ \circ } + { 360 }^{ \circ } k \qquad x = { 75 }^{ \circ } + { 180 }^{ \circ }k \end{matrix}\right

 

7.2 \cos { x } = 3 \tan { x }

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TRIGONOMETRIC IDENTITIES USED:

\tan { x } = \frac { \sin { x } }{ \cos { x } }

\cos^{ 2 }{ x } = 1 - \sin^{ 2 }{ x }

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2 \cos { x } = \frac { 3 \sin { x } }{ \cos { x } }

2 \cos^{ 2 }{ x } = 3 \sin { x }

2(1 - \sin^{ 2 }{ x }) = 3 \sin { x }

2 - 2\sin^{ 2 }{ x } - 3 \sin { x }

Let \sin { x } = y

2 - 2 { y }^{ 2 } - 3y = 0

2 { y }^{ 2 } + 3y - 2 =0

2 { y }^{ 2 } + y(4 - 1) - 2 =0

2 { y }^{ 2 } + 4y - y - 2 =0

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

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

y = \frac { 1 }{ 2 } \qquad y = -2

\sin { x } = \frac { 1 }{ 2 } \qquad \sin { x } = -2

\left\{\begin{matrix} { x }_{ 1 } = { 30 }^{ \circ } + { 360 }^{ \circ } k\\ { x }_{ 2 } = { 150 }^{ \circ } + { 360 }^{ \circ } k \end{matrix}\right

There is no solution of \sin { x } = -2 because -1 \leq \sin { x } \leq 1

 

8.\sin { 2x } . \cos { x } = 6 \sin^{ 3 }{ x }

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TRIGONOMETRIC IDENTITIES USED:

\sin { 2x } = 2 \sin { x } \cos { x }

\tan { x } = \frac { \sin { x } }{ \cos { x } }

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2 \sin { x } . \cos { x } . \cos { x } = 6 \sin^{ 3 }{ x }

\sin { x } (\cos^{ 2 }{ x } - 3 \sin^{ 2 }{ x }) = 0

\sin { x } = 0 \Rightarrow \left\{\begin{matrix} x = { 0 }^{ \circ } + { 360 }^{ \circ } k\\ x = { 180 }^{ \circ } + { 360 }^{ \circ } k \end{matrix}\right \Rightarrow x = { 0 }^{ \circ } + { 180 }^{ \circ } k

\cos^{ x } - 3 \sin^{ 2 }{ x } = 0

\cos^{ x }  = 3 \sin^{ 2 }{ x }

\frac { 1 }{ 3 } = \frac { \sin^{ 2 }{ x } }{ \cos^{ 2 }{ x } }

\tan^{ 2 }{ x } = \frac { 1 }{ 3 } \qquad \tan { x } = \pm \frac { \sqrt { 3 } }{ 3 }

\tan { x } = \frac { \sqrt { 3 } }{ 3 } \Rightarrow x = { 30 }^{ \circ } + { 180 }^{ \circ } k

\tan { x } = - \frac { \sqrt { 3 } }{ 3 } \Rightarrow x = { 150 }^{ \circ } + { 180 }^{ \circ } k

9.4 \sin { \frac { x }{ 2 } } + 2 \cos^{ 2 }{ x } = 3

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TRIGONOMETRIC IDENTITIES USED:

\cos^{ 2 }{ x } = \cos^{ 2 }{ x } - \sin^{ 2 }{ x }

\cos^{ 2 }{ x } + \sin^{ 2 }{ x } = 1 \qquad \cos^{ 2 }{ x } = 1 - \sin^{ 2 }{ x }

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4 \sin { \frac { x }{ 2 } } + 2 ( \cos^{ 2 }{ \frac { x }{ 2 } } - \sin^{ 2 }{ \frac { x }{ 2 } } ) = 3

4 \sin { \frac { x }{ 2 } } + 2 \cos^{ 2 }{ \frac { x }{ 2 } } - 2 \sin^{ 2 }{ \frac { x }{ 2 } } = 3

4 \sin { \frac { x }{ 2 } } + 2 (1 - \sin^{ 2 }{ \frac { x }{ 2 } }) - 2 \sin^{ 2 }{ \frac { x }{ 2 } } = 3

4 \sin { \frac { x }{ 2 } } + 2 - 2 \sin^{ 2 }{ \frac { x }{ 2 } } - 2 \sin^{ 2 }{ \frac { x }{ 2 } } = 3

4 \sin { \frac { x }{ 2 } } - 2 \sin^{ 2 }{ \frac { x }{ 2 } } - 2 \sin^{ 2 }{ \frac { x }{ 2 } } - 1 = 0

4 \sin^{ 2 }{ \frac { x }{ 2 } } - 4 \sin { \frac { x }{ 2 } } + 1 = 0

Let \sin { \frac { x }{ 2 } } = y:

4 { y }^{ 2 } - 4y + 1 = 0

{ (2y) }^{ 2 } - 2( 2y )( 1 ) + { (1) }^{ 2 } = 0

{ (2y - 1) }^{ 2 } = 0

2y - 1 = 0

2y = 1

y = \frac { 1 }{ 2 }

\sin { \frac { x }{ 2 } } = \frac { 1 }{ 2 } \Rightarrow \left\{\begin{matrix} \frac { x }{ 2 } = { 30 }^{ \circ } + { 360 }^{ \circ } k \qquad x = { 60 }^{ \circ } + { 360 }^{ \circ } k \\ \frac { x }{ 2 } = { 150 }^{ \circ } + { 360 }^{ \circ } k \qquad x = { 300 }^{ \circ } + { 360 }^{ \circ } k  \end{matrix}\right

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Hamza

Hi! I am Hamza and I am from Pakistan. My hobbies are reading, writing and playing chess. Currently, I am a student enrolled in the Chemical Engineering Bachelor program.