Exercise 1

Solve the following trigonometric equations:

1 \sin { x } = 0

2  \cos { x } = 0

3  \tan { x } = 0

4  \sin { x } = 1

5  \cos { x } = 1

6  \tan { x } = 1

7  \sin { x } = -1

8  \cos { x } = -1

9  \tan { x } = -1

10  \sin { x } = \frac { 1 }{ 2 }

11  \sin { x } = - \frac { 1 }{ 2 }

12  \cos { x } = \frac { 1 }{ 2 }

13  \cos { x } = - \frac { 1 }{ 2 }

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Exercise 2

\sin { (x + \frac { \pi }{ 4 }) } = \frac { \sqrt { 3 } }{ 2 }

Exercise 3

3 \sin^{ 2 }{ x } - 5 \sin { x } + 2 = 0

Exercise 4

\cos { 2x } = 1 + 4 \sin { x }

Exercise 5

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

Exercise 6

\cos { 8x } + \cos { 6x } = 2 \cos { { 210 }^{\circ } } . \cos { x }

Exercise 7

\tan { 2x } = - \tan { x }

Exercise 8

4 \sin { (x - { 30 }^{\circ }) } \cos { (x - { 30 }^{\circ }) } = \sqrt { 3 }

 

Solution of exercise 1

1 \sin { x } = 0

\arcsin { (\sin { x }) } = \arcsin { 0 } \qquad f . { f }^{ -1 } = x

x = \arcsin { 0 } \Rightarrow \left\{\begin{matrix} {x}_{1} = { 0 }^{\circ } + { 360 }^{\circ }k\\ {x}_{2} = { 180 }^{\circ } + { 360 }^{\circ }k \end{matrix}\right

{ x }_{ 1 } = { 0 }^{\circ }, { 360 }^{\circ }, { 720 }^{\circ }, ...

{ x }_{ 2 } = { 180 }^{\circ }, { 540 }^{\circ }, { 900 }^{\circ }, ...

x = { 0 }^{\circ } + { 180 }^{\circ } k

 

 

2  \cos { x } = 0

\arccos { (\cos { x }) } = \arccos { 0 } \qquad f . { f }^{ -1 } = x

x = \arccos { 0 } \Rightarrow \left\{\begin{matrix} {x}_{1} = { 90 }^{\circ } + { 360 }^{\circ }k\\ {x}_{2} = { 270 }^{\circ } + { 360 }^{\circ }k \end{matrix}\right

{ x }_{ 1 } = { 90 }^{\circ }, { 450 }^{\circ }, { 810 }^{\circ }, ...

{ x }_{ 2 } = { 270 }^{\circ }, { 630 }^{\circ }, { 990 }^{\circ }, ...

x = { 90 }^{\circ } + { 180 }^{\circ } k

 

 

3  \tan { x } = 0

\arctan { (\tan { x }) } = \arctan { 0 } \qquad f . { f }^{ -1 } = x

x = \arctan { 0 } \qquad x = { 0 }^{\circ } + { 180 }^{\circ } k

 

 

4  \sin { x } = 1

\arcsin { (\sin { x }) } = \arcsin { 1 } \qquad f . { f }^{ -1 } = x

x = \arcsin { 1 } \qquad x = { 90 }^{\circ } + { 360 }^{\circ } k

 

 

5  \cos { x } = 1

\arccos { (\cos { x }) } = \arccos { 1 } \qquad f . { f }^{ -1 } = x

x = \arccos { 1 } \qquad x = { 0 }^{\circ } + { 360 }^{\circ } k

 

 

6  \tan { x } = 1

\arctan { (\tan { x }) } = \arctan { 1 } \qquad f . { f }^{ -1 } = x

x = \arctan { 1 } \qquad x = { 45 }^{\circ } + { 180 }^{\circ } k

 

7  \sin { x } = -1

\arcsin { (\sin { x }) } = \arcsin { -1 } \qquad f . { f }^{ -1 } = x

x = \arcsin { (-1) } \qquad x = { 270 }^{\circ } + { 360 }^{\circ } k

 

 8 \cos { x } = -1

\arccos { (\cos { x }) } = \arccos { -1 } \qquad f . { f }^{ -1 } = x

x = \arccos { (-1) } \qquad x = { 180 }^{\circ } + { 360 }^{\circ } k

 

 

9  \tan { x } = -1

\arctan { (\tan { x }) } = \arctan { -1 } \qquad f . { f }^{ -1 } = x

x = \arctan { (-1) } \qquad x = { 135 }^{\circ } + { 180 }^{\circ } k

 

 

10  \sin { x } = \frac { 1 }{ 2 }

\arcsin { (\sin { x }) } = \arcsin { (\frac { 1 }{ 2 }) } \qquad f . { f }^{ -1 } = x

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

 

 

11  \sin { x } = - \frac { 1 }{ 2 }

\arcsin { (\sin { x }) } = \arcsin { (- \frac { 1 }{ 2 }) } \qquad f . { f }^{ -1 } = x

x = \arcsin { (-\frac { 1 }{ 2 }) } \Rightarrow \left\{\begin{matrix} {x}_{1} = { 30 }^{\circ } + { 360 }^{\circ }k\\ {x}_{2} = { 330 }^{\circ } + { 360 }^{\circ }k \end{matrix}\right

 

12  \cos { \frac { 1 }{ 2 }}

\arccos { (\cos { x }) } = \arccos { (\frac { 1 }{ 2 }) } \qquad f . { f }^{ -1 } = x

x = \arccos { (\frac { 1 }{ 2 }) } \Rightarrow \left\{\begin{matrix} {x}_{1} = { 60 }^{\circ } + { 360 }^{\circ }k\\ {x}_{2} = { 300 }^{\circ } + { 360 }^{\circ }k \end{matrix}\right

 

13 \cos { - \frac { 1 }{ 2 } }

\arccos { (\cos { x }) } = \arccos { (-\frac { 1 }{ 2 }) } \qquad f . { f }^{ -1 } = x

x = \arccos { (-\frac { 1 }{ 2 }) } \Rightarrow \left\{\begin{matrix} {x}_{1} = { 120 }^{\circ } + { 360 }^{\circ }k\\ {x}_{2} = { 240 }^{\circ } + { 360 }^{\circ }k \end{matrix}\right

 

Solution of exercise 2

\sin { (x + \frac { \pi }{ 4 }) } = \frac { \sqrt { 3 } }{ 2 }

\frac { \sqrt { 3 } }{ 2 } \rightarrow \left\{\begin{matrix} \sin { { 60 }^{\circ } } \\ \sin { { 120 }^{\circ } } \end{matrix}\right

It means that the value of \arcsin(\frac { \sqrt { 3 } }{ 2 }) will result in two values.

 

Converting \frac { \pi }{ 4 } to degrees:

\frac { \pi }{ 4 } \times \frac { 180 }{ \pi } = { 45 }^{\circ }

 

x + { 45 }^{\circ } = { 60 }^{\circ } \qquad x = { 15 }^{\circ }

{ x }_{ 1 } = { 15 }^{\circ } + { 360 }^{\circ }k

 

x + { 45 }^{\circ } = { 120 }^{\circ } \qquad x = { 15 }^{\circ }

{ x }_{ 2 } = { 75 }^{\circ } + { 360 }^{\circ }k

 

 

Solution of exercise 3

3 \sin^{ 2 }{ x } - 5 \sin { x } + 2 = 0

Let \sin { x } = y:

3 { y }^{ 2 } - 5 y + 2 = 0

y = \frac { - ( -5 ) \pm \sqrt { { (-5) }^{ 2 } - 4(3)(2) }}{ 2( 3 ) }

y = \frac { 5 \pm \sqrt {  25 - 24 } }{ 6 }

y = \frac { 5 \pm 1 }{ 6 }

y = \frac { 5 - 1 }{ 6 } \qquad y = \frac { 5 + 1 }{ 6 }

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

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

Since \sin { x } = y:

\sin { x } = \frac { 2 }{ 3 } \qquad \sin { x } = 1

For \sin { x } = 1 \qquad \qquad x = { 90 }^{\circ } + { 360 }^{\circ }k

For \sin { x } = \frac { 2 }{ 3 } \qquad \qquad x = \left\{\begin{matrix} { 41.81 }^{\circ } + { 360 }^{\circ }k \\ { 138.18 }^{\circ } + { 360 }^{\circ }k \end{matrix}\right

 

 

Solution of exercise 4

\cos { 2x } = 1 + 4 \sin { x }

______________________________________________________

TRIGONOMETRIC IDENTITIES USED:

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

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

______________________________________________________

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

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

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

2 \sin { x } ( \sin { x } + 2) = 0 \Rightarrow \{\begin{matrix} \sin { x } =0 \\ \sin { x } + 2 = 0 \end{matrix}\right

x = \arcsin { 0 } \Rightarrow \{\begin{matrix} {x}_{1} = { 0 }^{\circ } + { 360 }^{\circ }k\\ {x}_{2} = { 180 }^{\circ } + { 360 }^{\circ }k \end{matrix}\right

x = \arcsin { (-2) } \qquad No \quad Solution

 

 

Solution of exercise 5

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

______________________________________________________

TRIGONOMETRIC IDENTITIES USED:

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

______________________________________________________

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

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

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

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

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

 

Solution of exercise 6

\cos { 8x } + \cos { 6x } = 2 \cos { { 210 }^{\circ } } . \cos { x }

______________________________________________________

TRIGONOMETRIC IDENTITIES USED:

\cos { \alpha} + \cos { \beta } = 2 \cos { \frac { \alpha + \beta }{ 2 } } \cos { \frac { \alpha - \beta }{ 2 } }

______________________________________________________

2 \cos { \frac { 8x + 6x  }{ 2 } } \cos { \frac { 8x - 6x }{ 2 } } = 2 \cos { { 210 }^{\circ } } . \cos { x }

2 \cos { \frac { 14x  }{ 2 } } \cos { \frac { 2x }{ 2 } } = 2 \cos { { 210 }^{\circ } } . \cos { x }

2 \cos { 7x } \cos { x } = 2 \cos { { 210 }^{\circ } } . \cos { x }

\cos { x } (\cos { 7x } - \cos { { 210 }^{\circ } }) = 0

\cos { x } = 0 \qquad x = \left\{\begin{matrix} x = { 90 }^{\circ } + { 360 }^{\circ }k\\ x = { 270 }^{\circ } + { 360 }^{\circ }k \end{matrix}\right \qquad x = { 90 }^{\circ } + { 180 }^{\circ }k

\cos { 7x } - \cos { { 270 }^{\circ } } = 0

7x = { 210 }^{\circ }

x = { 30 }^{\circ }

 

 

Solution of exercise 7

\tan { 2x } = - \tan { x }

\frac { 2 \tan { x } }{ 1 - \tan^{ 2 }{ x } } = - \tan { x }

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

\tan { x } = 0 \qquad x = { 0 }^{\circ } + { 180 }^{\circ }k

\tan { x } = \pm \sqrt { 3 } \qquad left\{\begin{matrix} x = { 60 }^{\circ } + { 180 }^{\circ }k\\ x = { 120 }^{\circ } + { 180 }^{\circ }k \end{matrix}\right

 

Solution of exercise 8

4 \sin { (x - { 30 }^{\circ }) } \cos { (x - { 30 }^{\circ }) } = \sqrt { 3 }

2 [2 \sin { (x - { 30 }^{\circ }) \cos { (x - { 30 }^{\circ }) } }] = \sqrt { 3 }

\sin { 2(x - { 30 }^{\circ }) } = \sqrt { \sqrt { 3 } }{ 2 }

2(x - { 30 }^{\circ }) = { 60 }^{\circ } + { 360 }^{\circ }k \qquad \qquad x = { 60 }^{\circ } + { 180 }^{\circ }k

2(x - { 30 }^{\circ }) = { 120 }^{\circ } + { 360 }^{\circ }k \qquad \qquad x = { 90 }^{\circ } + { 180 }^{\circ }k

 

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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.