maths.free › Trigonometry › Trigonometric equations

Trigonometric equations

Reference angles and the infinitely many solutions.

sin x = ½ is true at π/6, at 5π/6, and again every full turn. Solve on one period from the unit circle, then add 2πn for every integer n. The graph shows why the solutions repeat.

Nhazi ahụ ejirila · sin(x) = 1/2

Trigonometry · Trigonometric-Equations

Solve sin(x) = 1/2

\sin{\left(x \right)} = \frac{1}{2}

Nzọụkwụ site n'ụdị

  1. \sin{\left(x \right)} = \frac{1}{2}

    Start from the equation as given.

  2. \sin{\left(x \right)} - \frac{1}{2} = 0

    Bring everything to one side.

  3. \sin{\left(x \right)} - \frac{1}{2} = 0

    Isolate the trig function, then read the reference angle off the unit circle. Solutions repeat every period.

  4. \left\{2 n \pi + \frac{\pi}{6}\; \middle|\; n \in \mathbb{Z}\right\} \cup \left\{2 n \pi + \frac{5 \pi}{6}\; \middle|\; n \in \mathbb{Z}\right\}

    General solution over the reals (n is any integer).

  5. x = \frac{\pi}{6} \approx 0.52360 ,\; x = \frac{5 \pi}{6} \approx 2.6180

    Solve for the variable.

Gosi nzaghachi
x = \frac{\pi}{6} ,\; x = \frac{5 \pi}{6}

Ugbu a ị Họrọ nsogbu, mọọbụ tinye mọọbụ depụta gị. Nzọụkwụ ọbụla, inyogo, azịza ahụ echekwara ruo mgbe ịjụla.

Kwụsị ọrụ gị

Akwụkwọ ndebanye n'efu na-egbakwunye ntọala n'ụdị ọmụmụ ihe ọbụla, ndepụta nke ihe ị ga-eme, nsogbu ị ga-ewepụ n'otu ebe, na onye nkụzi ị ga-ajụ banyere ihuakwụkwọ a. Akwụkwọ ndepụta ihe bụ nke ọbụla mepere maka onye ọbụla, ọbụla ma ọ bụ na ọ bụghị.

Akaụntụ Nbanye

Aịkọn ndị ahụ ejirila ebe a

Tinye ọbụla simbụl maka nkọwa zuru ezu, inyogo, na ihe ọbụla ọbụla na-egosi.

Olee otú: Trigonometric equations

  1. Start from the equation as given.
  2. Bring everything to one side.
  3. Isolate the trig function, then read the reference angle off the unit circle. Solutions repeat every period.
  4. General solution over the reals (n is any integer).
  5. Solve for the variable.

Ajụjụ ndị mmadụ jụrụ

Why radians instead of degrees?

A radian is the angle whose arc equals the radius, so it is a pure ratio rather than an arbitrary 1/360 of a turn. With radians the derivative of sin x is exactly cos x; with degrees a factor of π/180 appears everywhere.

Why does sin x = 1/2 have infinitely many solutions?

Sine repeats every full turn, and within one turn it reaches 1/2 twice (at π/6 and 5π/6). Add any whole number of turns to either and it is still true.

How do I remember the exact values?

Two triangles: the 45-45-90 with sides 1, 1, √2 and the 30-60-90 with sides 1, √3, 2. Every value for 30°, 45° and 60° is a ratio of those sides; symmetry gives the rest of the circle.

Oge Trigonometry

The unit circleTrigonometric identitiesDegrees and radiansRight-triangle trigonometry (SOH-CAH-TOA)Law of sines and law of cosinesGraphs of sine, cosine and tangentInverse trigonometric functions