Math  /  Trigonometry

QuestionSolve 2sinx+1=02 \sin x+1=0. Show solution graphically.

Studdy Solution
Graph the function y=2sinx+1 y = 2\sin x + 1 and indicate the solutions:
1. Plot the graph of y=2sinx+1 y = 2\sin x + 1 .
2. Identify where the graph intersects the line y=0 y = 0 .
3. Mark the solutions x=7π6 x = \frac{7\pi}{6} and x=11π6 x = \frac{11\pi}{6} on the graph.

The solutions to the equation 2sinx+1=0 2 \sin x + 1 = 0 within the interval [0,2π][0, 2\pi] are:
x=7π6,11π6 x = \frac{7\pi}{6}, \frac{11\pi}{6}

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