the number of solutions of sin x= -sqrt3/2 for x between 0 and 2pi
two
sin x < 0 in QIII, QIV
To find the number of solutions of the equation sin(x) = -√3/2 for x between 0 and 2π, we can follow these steps:
Step 1: Identify the value of sin(x) for which the equation is satisfied.
In this case, sin(x) = -√3/2.
Step 2: Identify the angles in the range between 0 and 2π that have the same value of sin(x).
For sin(x) = -√3/2, we need to find angles in the range between 0 and 2π that have the same sine value.
Step 3: Use the unit circle or trigonometric values to find these angles.
Looking at the unit circle, we can see that sin(x) = -√3/2 corresponds to two angles, which are 11π/6 and 7π/6.
Alternatively, we can use the symmetry of the sine function to find these angles. The sine function is negative in the third and fourth quadrants, and sin(x) = -√3/2 is the same as sin(x) = √3/2 in the first and second quadrants. The angles in the first and second quadrants that satisfy sin(x) = √3/2 are π/3 and 2π/3. Therefore, the angles in the third and fourth quadrants are 4π/3 and 5π/3.
Step 4: Count the number of solutions within the given range.
Between 0 and 2π, we have 4 angles that satisfy sin(x) = -√3/2: π/3, 2π/3, 4π/3, and 5π/3.
Therefore, the number of solutions of sin(x) = -√3/2 for x between 0 and 2π is 4.