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Difficulty: MediumTrigonometric Graphs and Simple Equations

Which of the following sets contains all values of xx in the interval 0∘≤x≤360∘0^\circ \le x \le 360^\circ that satisfy the trigonometric equation 3sin⁡x+cos⁡x=0\sqrt{3}\sin x + \cos x = 0?

  1. 150∘ and 330∘150^\circ \text{ and } 330^\circAnswer
  2. B
    30∘ and 210∘30^\circ \text{ and } 210^\circ
  3. C
    120∘ and 300∘120^\circ \text{ and } 300^\circ
  4. D
    150∘ only150^\circ \text{ only}

Answer

150∘ and 330∘150^\circ \text{ and } 330^\circ
The given equation 3sin⁡x+cos⁡x=0\sqrt{3}\sin x + \cos x = 0 simplifies to tan⁡x=−13\tan x = -\frac{1}{\sqrt{3}}. Since tangent is negative in the second and fourth quadrants with a reference angle of 30∘30^\circ, the solutions in the domain 0∘≤x≤360∘0^\circ \le x \le 360^\circ are 180∘−30∘=150∘180^\circ - 30^\circ = 150^\circ and 360∘−30∘=330∘360^\circ - 30^\circ = 330^\circ.

Step-by-Step Solution

1
Rearrange the trigonometric equation into single ratio form
3sin⁡x=−cos⁡x  ⟹  sin⁡xcos⁡x=−13  ⟹  tan⁡x=−13\sqrt{3}\sin x = -\cos x \implies \frac{\sin x}{\cos x} = -\frac{1}{\sqrt{3}} \implies \tan x = -\frac{1}{\sqrt{3}}
Dividing both sides by cos⁡x\cos x converts the sum of sine and cosine terms into a simple tangent equation.
2
Determine the reference angle
Reference angle α=30∘\text{Reference angle } \alpha = 30^\circ
The acute angle whose tangent is 13\frac{1}{\sqrt{3}} is 30∘30^\circ.
3
Identify the quadrants and find all solutions in 0∘≤x≤360∘0^\circ \le x \le 360^\circ
x=180∘−30∘=150∘x = 180^\circ - 30^\circ = 150^\circ (Quadrant II) and x=360∘−30∘=330∘x = 360^\circ - 30^\circ = 330^\circ (Quadrant IV)
The tangent function is negative in Quadrants II and IV.

Key Concept

Solving simple trigonometric equations by reducing to a basic ratio and finding all solutions within a given domain.
Estimated Time:1m 30s
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Topics

Trigonometric Graphs and Simple Equations

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