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.
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![{\displaystyle n}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a601995d55609f2d9f5e233e36fbe9ea26011b3b) |
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The equation
has solutions only when
is within the interval
. If
is within this interval, then we first find an
such that:
![{\displaystyle \alpha =\arcsin(n)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eb9c815f6b479ae6b1559a4345539d11f3417e04)
The solutions are then:
![{\displaystyle x=\alpha +2k\pi }](https://wikimedia.org/api/rest_v1/media/math/render/svg/d2d790c01256d391e2efe73e24920165fe9e204f)
![{\displaystyle x=\pi -\alpha +2k\pi }](https://wikimedia.org/api/rest_v1/media/math/render/svg/92548a0abdb567c86f89bf67834f1633e156cd86)
Where
is an integer.
In the cases when
equals 1, 0 or -1 these solutions have simpler forms which are summarized in the table on the right.
For example, to solve:
![{\displaystyle \sin {\bigl (}{\tfrac {x}{2}}{\bigr )}={\frac {\sqrt {3}}{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fa38784ca99ae06e19754c0223ff283372dcd2bc)
First find
:
![{\displaystyle \alpha =\arcsin {\bigl (}{\tfrac {\sqrt {3}}{2}}{\bigr )}={\frac {\pi }{3}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6ca712b83677b234f07a5b086fe0264acf08dbdb)
Then substitute in the formulae above:
![{\displaystyle {\frac {x}{2}}={\frac {\pi }{3}}+2k\pi }](https://wikimedia.org/api/rest_v1/media/math/render/svg/76e18efd5638ed03c06b22a8284034988f441795)
![{\displaystyle {\frac {x}{2}}=\pi -{\frac {\pi }{3}}+2k\pi }](https://wikimedia.org/api/rest_v1/media/math/render/svg/f31e2020b0bdfc19e45ea97e4a4bf95c9fb45137)
Solving these linear equations for
gives the final answer:
![{\displaystyle x={\frac {2\pi }{3}}(1+6k)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/55365112eccc6f381c95a9d4aa1d09fc62789f7d)
![{\displaystyle x={\frac {4\pi }{3}}(1+3k)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/390b6c74fa56df6bc7a711d1cadab3d2e308dacd)
Where
is an integer.
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![{\displaystyle n}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a601995d55609f2d9f5e233e36fbe9ea26011b3b) |
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Like the sine equation, an equation of the form
only has solutions when n is in the interval
. To solve such an equation we first find one angle
such that:
![{\displaystyle \alpha =\arccos(n)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/50f3081b42574a05ed532a47c4f47a9eed830572)
Then the solutions for
are:
![{\displaystyle x=\pm \alpha +2k\pi }](https://wikimedia.org/api/rest_v1/media/math/render/svg/cf6e6741fa9785d99860ce93cb1a4c363f5e3537)
Where
is an integer.
Simpler cases with
equal to 1, 0 or -1 are summarized in the table on the right.
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![{\displaystyle n!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bae971720be3cc9b8d82f4cdac89cb89877514a6) |
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General case
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An equation of the form
has solutions for any real
. To find them we must first find an angle
such that:
![{\displaystyle \alpha =\arctan(n)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3909c5d17168211c2a977158e4a603cf851e8d6c)
After finding
, the solutions for
are:
![{\displaystyle x=\alpha +k\pi }](https://wikimedia.org/api/rest_v1/media/math/render/svg/55c937413f0754206148f08ead34a2fda8ae027c)
When
equals 1, 0 or -1 the solutions have simpler forms which are shown in the table on the right.
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![{\displaystyle n}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a601995d55609f2d9f5e233e36fbe9ea26011b3b) |
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General case
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The equation
has solutions for any real
. To find them we must first find an angle
such that:
![{\displaystyle \alpha =\operatorname {arccot}(n)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/626d41813ce34326226ab5e2a0877f952f907b0f)
After finding
, the solutions for
are:
![{\displaystyle x=\alpha +k\pi }](https://wikimedia.org/api/rest_v1/media/math/render/svg/55c937413f0754206148f08ead34a2fda8ae027c)
When
equals 1, 0 or -1 the solutions have simpler forms which are shown in the table on the right.
The trigonometric equations
and
can be solved by transforming them to other basic equations:
![{\displaystyle \csc(x)=n\ \Leftrightarrow \ {\frac {1}{\sin(x)}}=n\ \Leftrightarrow \ \sin(x)={\frac {1}{n}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aaf64fa9ee8fda96053d9b106214af8cfa93e1e7)
![{\displaystyle \sec(x)=n\ \Leftrightarrow \ {\frac {1}{\cos(x)}}=n\ \Leftrightarrow \ \cos(x)={\frac {1}{n}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5859646e6a2c3c8b6382967d05ed6d5fd81ac336)
Generally, to solve trigonometric equations we must first transform them to a basic trigonometric equation using the trigonometric identities. This sections lists some common examples.
To solve this equation we will use the identity:
![{\displaystyle a\sin(x)+b\cos(x)={\sqrt {a^{2}+b^{2}}}\sin(x+\alpha )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fa557ce90802439ebb056813064257ee085e5344)
![{\displaystyle \alpha ={\begin{cases}\arctan {\bigl (}{\frac {b}{a}}{\bigr )},&{\mbox{if }}a>0\\\pi +\arctan {\bigl (}{\frac {b}{a}}{\bigr )},&{\mbox{if }}a<0\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/38d9810de236e31bee99757ccfce555d0949f0f3)
The equation becomes:
![{\displaystyle {\sqrt {a^{2}+b^{2}}}\sin(x+\alpha )=c}](https://wikimedia.org/api/rest_v1/media/math/render/svg/889a161ffa821c671ba297f489cf87efe84d1154)
![{\displaystyle \sin(x+\alpha )={\frac {c}{\sqrt {a^{2}+b^{2}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8995346751fdb804f5dda11bbf9edb43631007d8)
This equation is of the form
and can be solved with the formulae given above.
For example we will solve:
![{\displaystyle \sin(3x)-{\sqrt {3}}\cos(3x)=-{\sqrt {3}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e0380d9f940ae24b1665bfad6128a4b42d08e604)
In this case we have:
![{\displaystyle a=1,b=-{\sqrt {3}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/be195c7ca4529deb195770914cfa7589f3eebdb7)
![{\displaystyle {\sqrt {a^{2}+b^{2}}}={\sqrt {1^{2}+{\Big (}-{\sqrt {3}}{\Big )}^{2}}}=2}](https://wikimedia.org/api/rest_v1/media/math/render/svg/234a3eaef1679b96244cf1382e0f6abb944acba3)
![{\displaystyle \alpha =\arctan {\Big (}-{\sqrt {3}}{\Big )}=-{\frac {\pi }{3}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/94af1fa0f50a7fa90cabf17cc111ee6fdea5ad27)
Apply the identity:
![{\displaystyle 2\sin \left(3x-{\frac {\pi }{3}}\right)=-{\sqrt {3}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ed2666b354661cab89cf1d40546267e8af4007d2)
![{\displaystyle \sin \left(3x-{\frac {\pi }{3}}\right)=-{\frac {\sqrt {3}}{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c5281ca5bf7e62e19aa06e5d98a3dd9b3f9eb42a)
So using the formulae for
the solutions to the equation are:
![{\displaystyle 3x-{\frac {\pi }{3}}=-{\frac {\pi }{3}}+2k\pi \ \Leftrightarrow \ x={\frac {2k\pi }{3}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/59762eefc08654ee01fa01cf8d0c83d0e1720bd3)
![{\displaystyle 3x-{\frac {\pi }{3}}=\pi +{\frac {\pi }{3}}+2k\pi \ \Leftrightarrow \ x={\frac {\pi }{9}}(6k+5)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b666d8e1bfce189e69c442f2d23a3be298f36669)
Where
is an integer.