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
sin(x ) = n
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} has solutions only when n {\displaystyle n} is within the interval [ − 1 , 1 ] {\displaystyle [-1,1]} . If n {\displaystyle n} is within this interval, then we first find an α {\displaystyle \alpha } such that:
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} is an integer.
In the cases when n {\displaystyle n} equals 1, 0 or -1 these solutions have simpler forms which are summarized in the table on the right.
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} gives the final answer:
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} is an integer.
cos(x ) = n
edit
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} only has solutions when n is in the interval [ − 1 , 1 ] {\displaystyle [-1,1]} . To solve such an equation we first find one angle α {\displaystyle \alpha } such that:
α = arccos ( n ) {\displaystyle \alpha =\arccos(n)} Then the solutions for x {\displaystyle x} are:
x = ± α + 2 k π {\displaystyle x=\pm \alpha +2k\pi } Where k {\displaystyle k} is an integer.
Simpler cases with n {\displaystyle n} equal to 1, 0 or -1 are summarized in the table on the right.
tan(x ) = n
edit
n ! {\displaystyle n!}
tan ( x ) = n {\displaystyle \tan(x)=n}
General case
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} has solutions for any real n {\displaystyle n} . To find them we must first find an angle α {\displaystyle \alpha } such that:
α = arctan ( n ) {\displaystyle \alpha =\arctan(n)} After finding α {\displaystyle \alpha } , the solutions for x {\displaystyle x} are:
x = α + k π {\displaystyle x=\alpha +k\pi } When n {\displaystyle n} equals 1, 0 or -1 the solutions have simpler forms which are shown in the table on the right.
cot(x ) = n
edit
n {\displaystyle n}
cot ( x ) = n {\displaystyle \cot(x)=n}
General case
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} has solutions for any real n {\displaystyle n} . To find them we must first find an angle α {\displaystyle \alpha } such that:
α = arccot ( n ) {\displaystyle \alpha =\operatorname {arccot}(n)} After finding α {\displaystyle \alpha } , the solutions for x {\displaystyle x} are:
x = α + k π {\displaystyle x=\alpha +k\pi } When n {\displaystyle n} equals 1, 0 or -1 the solutions have simpler forms which are shown in the table on the right.
csc(x ) = n and sec(x ) = n
edit
The trigonometric equations csc ( x ) = n {\displaystyle \csc(x)=n} and sec ( x ) = n {\displaystyle \sec(x)=n} can be solved by transforming them to other basic equations:
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}}} Further examples
edit