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11. Trigonometric Equations
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Q21 of 142 Page 11

Mark the Correct alternative in the following:

The number of values of x in the interval [0. 5 π] satisfying the equation 3 sin2 x – 7 sin x + 2 = 0 is


3sin2x-7sin x+2=0


Solving the equation, we get




=19.47122


x= nπ + (-1)n a


For


n=0, x=a


n=1, x = π – a


n=2, x = 2π + a


n=3, x = 3π – a


n=4, x = 4π + a


n=5, x = 5π + a


So, there are 6 values less then 5π.


Option C.

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Questions · 142
11. Trigonometric Equations
1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 5 5 5 5 5 5 6 6 6 6 6 6 6 6 6 6 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 8 8 9 9 10 10 1 2 3 4 5 6 7 8 9 10 11 12 13 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21
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