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Trigonometric equationsTrigonometryHigh schoolabout 2:06

Find arccos(-√3/2) and every angle in [0°, 360°) with cos x = -√3/2: animated lesson

A narrated, step-by-step trigonometric equations video lesson with captions. Final answer: arccos(-√3/2) = 150° (5π/6); solutions in [0°, 360°): 150°, 210°; general: x = ±150° + 360°n.

Video with a synthetic English voice and captions. Pick another language for subtitles in that language, or open the interactive version.

Video narrated by a synthetic voice (Piper TTS, en_US-ljspeech-high voice (LJ Speech dataset, public domain)); the steps come from our maths engine and the answer was checked independently. Download MP4 (0.8 MB) · Captions (WebVTT)

Steps on screen

  1. 1. The problem
    Worked example
    Find arccos(-√3/2) and every angle in [0°, 360°) with cos x = -√3/2.
  2. 2. Step by step
    1. arccos⁡(−3/2)=150∘=5π6\operatorname{arccos}\left(-\sqrt{3}/2\right) = 150^\circ = \frac{5\pi}{6}
      Principal value of arccos, taken in the range [0, π]
    2. cos⁡(−x)=cos⁡x\cos(-x) = \cos x
      Cosine is even, so x and −x (that is, 360° − x) give the same value
    3. x=150∘, 210∘(5π6, 7π6)x = 150^\circ,\ 210^\circ \quad (\frac{5\pi}{6},\ \frac{7\pi}{6})
      Solutions in one full turn, 0° ≤ x < 360°
    4. x=±150∘+360∘nx = \pm 150^\circ + 360^\circ n
      General solution, n any integer
  3. 3. The answer
    arccos⁡(−3/2)=150∘=5π6,x=±150∘+360∘n\operatorname{arccos}\left(-\sqrt{3}/2\right) = 150^\circ = \frac{5\pi}{6},\quad x = \pm 150^\circ + 360^\circ n

    arccos(-√3/2) = 150°

  4. 4. Check
    • Substituting each solution back: cos(150°) ≈ -0.8660254038; cos(210°) ≈ -0.8660254038.

Transcript

The problem. Here is the problem: Find arccos(negative the square root of 3 over 2 and every angle in [0°, 360°) with cos x equals negative the square root of 3 over 2. Find the angle whose cos is negative the square root of 3 over 2: the principal value of arccos, every solution of cos(x) equals negative the square root of 3 over 2 in one turn, and the general solution.

Step by step. Principal value of arccos, taken in the range [0, pi]. 150° is a special angle, so the answer is exact. Cosine is even, so x and negative x (that is, 360° minus x) give the same value. Solutions in one full turn, 0° less than or equal to x less than 360°. General solution, n any integer.

The answer. So the answer is arccos(negative the square root of 3 over 2 equals 150° (5pi/6); solutions in [0°, 360°): 150°, 210°; general: x equals plus or minus 150° plus 360°n. An inverse trigonometric function returns one angle, the principal value. Because the trigonometric functions repeat, the equation cos(x) equals negative the square root of 3 over 2 has infinitely many solutions; the calculator lists those in one turn and the general pattern.

Check. Substituting each solution back: cos(150°) approximately negative 0.8660254038; cos(210°) approximately negative 0.8660254038.

About trigonometric equations

Solve trigonometric equations and give every solution.

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