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Solve the equation 2sin(x) + √3 = 0 for x in the interval [0, 2π].
Solve the equation 2sin(x) + √3 = 0 for x in the interval [0, 2π].
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Practice Questions
1 question
Q1
Solve the equation 2sin(x) + √3 = 0 for x in the interval [0, 2π].
5π/3
π/3
2π/3
4π/3
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Rearranging gives sin(x) = -√3/2, so x = 4π/3 and x = 5π/3.
Questions & Step-by-step Solutions
1 item
Q
Q: Solve the equation 2sin(x) + √3 = 0 for x in the interval [0, 2π].
Solution:
Rearranging gives sin(x) = -√3/2, so x = 4π/3 and x = 5π/3.
Steps: 6
Show Steps
Step 1: Start with the equation 2sin(x) + √3 = 0.
Step 2: Subtract √3 from both sides to isolate the term with sin(x): 2sin(x) = -√3.
Step 3: Divide both sides by 2 to solve for sin(x): sin(x) = -√3/2.
Step 4: Identify the angles where sin(x) equals -√3/2. These angles are in the third and fourth quadrants.
Step 5: The angles that satisfy sin(x) = -√3/2 are x = 4π/3 and x = 5π/3.
Step 6: Write the final solutions: x = 4π/3 and x = 5π/3.
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