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Difficulty: MediumTrigonometric Graphs and Simple Equations

For the domain 0∘≤x≤360∘0^\circ \le x \le 360^\circ, solve the trigonometric equation 2sin⁡x+3=02\sin x + \sqrt{3} = 0. Which of the following gives the complete set of values for xx?

  1. A
    60∘60^\circ and 120∘120^\circ
  2. B
    120∘120^\circ and 240∘240^\circ
  3. 240∘240^\circ and 300∘300^\circAnswer
  4. D
    210∘210^\circ and 330∘330^\circ

Answer

x=240∘x = 240^\circ and x=300∘x = 300^\circ
Rearranging 2sin⁡x+3=02\sin x + \sqrt{3} = 0 gives sin⁡x=−32\sin x = -\frac{\sqrt{3}}{2}. The reference angle for which sine equals 32\frac{\sqrt{3}}{2} is 60∘60^\circ. Since sine is negative in the third and fourth quadrants, the solutions are 180∘+60∘=240∘180^\circ + 60^\circ = 240^\circ and 360∘−60∘=300∘360^\circ - 60^\circ = 300^\circ.

Step-by-Step Solution

1
Isolate the trigonometric function in the equation
sin⁡x=−32\sin x = -\frac{\sqrt{3}}{2}
Subtract 3\sqrt{3} from both sides and divide by 22.
2
Find the basic reference angle
Reference angle α=60∘\alpha = 60^\circ
sin⁡(60∘)=32\sin(60^\circ) = \frac{\sqrt{3}}{2}.
3
Determine the relevant quadrants
Quadrants III and IV
The sine function is negative in the third and fourth quadrants.
4
Calculate the solutions within the interval 0∘≤x≤360∘0^\circ \le x \le 360^\circ
Quadrant III: x=180∘+60∘=240∘x = 180^\circ + 60^\circ = 240^\circ; Quadrant IV: x=360∘−60∘=300∘x = 360^\circ - 60^\circ = 300^\circ
Apply quadrant reduction formulas for Quadrant III (180∘+α180^\circ + \alpha) and Quadrant IV (360∘−α360^\circ - \alpha).

Key Concept

Solving simple trigonometric equations using reference angles and quadrant rules
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Topics

Trigonometric Graphs and Simple Equations

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