Question

The number of θ values of which satisfy the equation 6sin3θ + 25sin2θ − 39sinθ − 70 = 0 is

Answer 1

Given: The equation 6sin³θ + 25sin²θ − 39sinθ − 70 = 0

To find: The number of θ values.

Solution:

• Now we have provided with the equation:

                     6sin³θ + 25sin²θ − 39sinθ − 70 = 0

• The roots of the equations can be:

                     ( sinθ − 2) x ( sinθ + 5 ) x ( 6sinθ + 7 ) = 0

• From this, we get:

                     sinθ = 2, -5 and 6/7

• But we know the range of sinθ is from -1 to 1. So:

                     sinθ ≠ 2, -5

• Then sinθ = 6/7

• So, θ = 2nπ - arc sin 7/6

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Answer:

             So the value of θ will be 2nπ - arc sin 7/6