Trigonometric equations are equations including trigonometric functions. If they have only such functions and constants, then the solution involves finding an unknown which is an argument to a trigonometric function.
Basic trigonometric equations
edit
n
{\displaystyle n}
sin
(
x
)
=
n
{\displaystyle \sin(x)=n}
|
n
|
<
1
{\displaystyle |n|<1}
x
=
α
+
2
k
π
x
=
π
−
α
+
2
k
π
α
∈
[
−
π
2
,
π
2
]
{\displaystyle {\begin{matrix}x=\alpha +2k\pi \\x=\pi -\alpha +2k\pi \\\alpha \in \left[-{\frac {\pi }{2}},{\frac {\pi }{2}}\right]\end{matrix}}}
n
=
−
1
{\displaystyle n=-1}
x
=
−
π
2
+
2
k
π
{\displaystyle x=-{\begin{matrix}{\frac {\pi }{2}}\end{matrix}}+2k\pi }
n
=
0
{\displaystyle n=0}
x
=
k
π
{\displaystyle x=k\pi }
n
=
1
{\displaystyle n=1}
x
=
π
2
+
2
k
π
{\displaystyle x={\begin{matrix}{\frac {\pi }{2}}\end{matrix}}+2k\pi }
|
n
|
>
1
{\displaystyle |n|>1}
x
∈
∅
{\displaystyle x\in \varnothing }
The equation
sin
(
x
)
=
n
{\displaystyle \sin(x)=n}
n
{\displaystyle n}
[
−
1
,
1
]
{\displaystyle [-1,1]}
n
{\displaystyle n}
α
{\displaystyle \alpha }
α
=
arcsin
(
n
)
{\displaystyle \alpha =\arcsin(n)}
The solutions are then:
x
=
α
+
2
k
π
{\displaystyle x=\alpha +2k\pi }
x
=
π
−
α
+
2
k
π
{\displaystyle x=\pi -\alpha +2k\pi }
Where
k
{\displaystyle k}
In the cases when
n
{\displaystyle n}
For example, to solve:
sin
(
x
2
)
=
3
2
{\displaystyle \sin {\bigl (}{\tfrac {x}{2}}{\bigr )}={\frac {\sqrt {3}}{2}}}
First find
α
{\displaystyle \alpha }
α
=
arcsin
(
3
2
)
=
π
3
{\displaystyle \alpha =\arcsin {\bigl (}{\tfrac {\sqrt {3}}{2}}{\bigr )}={\frac {\pi }{3}}}
Then substitute in the formulae above:
x
2
=
π
3
+
2
k
π
{\displaystyle {\frac {x}{2}}={\frac {\pi }{3}}+2k\pi }
x
2
=
π
−
π
3
+
2
k
π
{\displaystyle {\frac {x}{2}}=\pi -{\frac {\pi }{3}}+2k\pi }
Solving these linear equations for
x
{\displaystyle x}
x
=
2
π
3
(
1
+
6
k
)
{\displaystyle x={\frac {2\pi }{3}}(1+6k)}
x
=
4
π
3
(
1
+
3
k
)
{\displaystyle x={\frac {4\pi }{3}}(1+3k)}
Where
k
{\displaystyle k}
n
{\displaystyle n}
cos
(
x
)
=
n
{\displaystyle \cos(x)=n}
|
n
|
<
1
{\displaystyle |n|<1}
x
=
±
α
+
2
k
π
α
∈
[
0
,
π
]
{\displaystyle {\begin{matrix}x=\pm \alpha +2k\pi \\\alpha \in [0,\pi ]\end{matrix}}}
n
=
−
1
{\displaystyle n=-1}
x
=
π
+
2
k
π
{\displaystyle x=\pi +2k\pi }
n
=
0
{\displaystyle n=0}
x
=
π
2
+
k
π
{\displaystyle x={\begin{matrix}{\frac {\pi }{2}}\end{matrix}}+k\pi }
n
=
1
{\displaystyle n=1}
x
=
2
k
π
{\displaystyle x=2k\pi }
|
n
|
>
1
{\displaystyle |n|>1}
x
∈
∅
{\displaystyle x\in \varnothing }
Like the sine equation, an equation of the form
cos
(
x
)
=
n
{\displaystyle \cos(x)=n}
[
−
1
,
1
]
{\displaystyle [-1,1]}
α
{\displaystyle \alpha }
α
=
arccos
(
n
)
{\displaystyle \alpha =\arccos(n)}
Then the solutions for
x
{\displaystyle x}
x
=
±
α
+
2
k
π
{\displaystyle x=\pm \alpha +2k\pi }
Where
k
{\displaystyle k}
Simpler cases with
n
{\displaystyle n}
n
{\displaystyle n}
tan
(
x
)
=
n
{\displaystyle \tan(x)=n}
General
x
=
α
+
k
π
α
∈
[
−
π
2
,
π
2
]
{\displaystyle {\begin{matrix}x=\alpha +k\pi \\\alpha \in \left[-{\frac {\pi }{2}},{\frac {\pi }{2}}\right]\end{matrix}}}
n
=
−
1
{\displaystyle n=-1}
x
=
−
π
4
+
k
π
{\displaystyle x=-{\begin{matrix}{\frac {\pi }{4}}\end{matrix}}+k\pi }
n
=
0
{\displaystyle n=0}
x
=
k
π
{\displaystyle x=k\pi }
n
=
1
{\displaystyle n=1}
x
=
π
4
+
k
π
{\displaystyle x={\begin{matrix}{\frac {\pi }{4}}\end{matrix}}+k\pi }
An equation of the form
tan
(
x
)
=
n
{\displaystyle \tan(x)=n}
n
{\displaystyle n}
α
{\displaystyle \alpha }
α
=
arctan
(
n
)
{\displaystyle \alpha =\arctan(n)}
After finding
α
{\displaystyle \alpha }
x
{\displaystyle x}
x
=
α
+
k
π
{\displaystyle x=\alpha +k\pi }
When
n
{\displaystyle n}
n
{\displaystyle n}
cot
(
x
)
=
n
{\displaystyle \cot(x)=n}
General
x
=
α
+
k
π
α
∈
[
0
;
π
]
{\displaystyle {\begin{matrix}x=\alpha +k\pi \\\alpha \in \left[0;\pi \right]\end{matrix}}}
n
=
−
1
{\displaystyle n=-1}
x
=
−
3
π
4
+
k
π
{\displaystyle x=-{\begin{matrix}{\frac {3\pi }{4}}\end{matrix}}+k\pi }
n
=
0
{\displaystyle n=0}
x
=
π
2
+
k
π
{\displaystyle x={\begin{matrix}{\frac {\pi }{2}}\end{matrix}}+k\pi }
n
=
1
{\displaystyle n=1}
x
=
π
4
+
k
π
{\displaystyle x={\begin{matrix}{\frac {\pi }{4}}\end{matrix}}+k\pi }
The equation
cot
(
x
)
=
n
{\displaystyle \cot(x)=n}
n
{\displaystyle n}
α
{\displaystyle \alpha }
α
=
arccot
(
n
)
{\displaystyle \alpha =\operatorname {arccot}(n)}
After finding
α
{\displaystyle \alpha }
x
{\displaystyle x}
x
=
α
+
k
π
{\displaystyle x=\alpha +k\pi }
When
n
{\displaystyle n}
x ) = n and sec(x ) = n
edit
The trigonometric equations
csc
(
x
)
=
n
{\displaystyle \csc(x)=n}
sec
(
x
)
=
n
{\displaystyle \sec(x)=n}
csc
(
x
)
=
n
⇔
1
sin
(
x
)
=
n
⇔
sin
(
x
)
=
1
n
{\displaystyle \csc(x)=n\ \Leftrightarrow \ {\frac {1}{\sin(x)}}=n\ \Leftrightarrow \ \sin(x)={\frac {1}{n}}}
sec
(
x
)
=
n
⇔
1
cos
(
x
)
=
n
⇔
cos
(
x
)
=
1
n
{\displaystyle \sec(x)=n\ \Leftrightarrow \ {\frac {1}{\cos(x)}}=n\ \Leftrightarrow \ \cos(x)={\frac {1}{n}}}
![{\displaystyle {\begin{matrix}x=\alpha +2k\pi \\x=\pi -\alpha +2k\pi \\\alpha \in \left[-{\frac {\pi }{2}},{\frac {\pi }{2}}\right]\end{matrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d1a3e9f50511c58a1b5da92a54ebaa528586f4b0)
![{\displaystyle [-1,1]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/51e3b7f14a6f70e614728c583409a0b9a8b9de01)
![{\displaystyle {\begin{matrix}x=\pm \alpha +2k\pi \\\alpha \in [0,\pi ]\end{matrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eb327bcece984447ffb90d738e9f5c2246d7375d)
![{\displaystyle {\begin{matrix}x=\alpha +k\pi \\\alpha \in \left[-{\frac {\pi }{2}},{\frac {\pi }{2}}\right]\end{matrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/091023c5ebe89e390dc7b3c08dc2aafd52f739cb)
![{\displaystyle {\begin{matrix}x=\alpha +k\pi \\\alpha \in \left[0;\pi \right]\end{matrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a09654bc73208cf0114b0bbebf26b68271c20ad0)
