Verify that 7pi/6 is a solution to
2sin2x–sinx=1.
To verify if 7π/6 is a solution to the equation 2sin2x – sinx = 1, we need to substitute this value into the equation and see if both sides are equal.
Step 1: Start with the given equation:
2sin2x – sinx = 1.
Step 2: Substitute x = 7π/6 into the equation:
2sin(2(7π/6)) – sin(7π/6) = 1.
Step 3: Simplify the expression inside the sine functions:
2sin(14π/6) – sin(7π/6) = 1.
Step 4: Convert the angles to a common denominator:
2sin(7π/3) – sin(7π/6) = 1.
Step 5: Recall the unit circle and the values of sine at these angles:
sin(7π/3) = sin(π/3) = √3/2,
sin(7π/6) = sin(π/6) = 1/2.
Step 6: Substitute these values back into the equation:
2(√3/2) – (1/2) = 1,
√3 – 1 = 1.
Step 7: Simplify the expression:
√3 – 1 = 1,
√3 = 2.
Step 8: Since this equation is not true, the value 7π/6 is not a solution to the equation 2sin2x – sinx = 1.
Therefore, 7π/6 is not a solution to the given equation.