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How do you solve sin x equals 6? - Answers

First, take the inverse sine of both sides of the equation. That gives you x = sin-1(6), which is sadly undefined...in reality, but who needs that! It can be proven that sin-1(x) = -i*log[i*x + √(1-x2)] So in this case: = -i*log[i*6 + √(1-36)] = -i*log[6*i + √(-35)] = -i*log(11.916*i) = 1.57 - 2.48*i



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How do you solve sin x equals 6? - Answers

https://math.answers.com/calculus/How_do_you_solve_sin_x_equals_6

First, take the inverse sine of both sides of the equation. That gives you x = sin-1(6), which is sadly undefined...in reality, but who needs that! It can be proven that sin-1(x) = -i*log[i*x + √(1-x2)] So in this case: = -i*log[i*6 + √(1-36)] = -i*log[6*i + √(-35)] = -i*log(11.916*i) = 1.57 - 2.48*i



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https://math.answers.com/calculus/How_do_you_solve_sin_x_equals_6

How do you solve sin x equals 6? - Answers

First, take the inverse sine of both sides of the equation. That gives you x = sin-1(6), which is sadly undefined...in reality, but who needs that! It can be proven that sin-1(x) = -i*log[i*x + √(1-x2)] So in this case: = -i*log[i*6 + √(1-36)] = -i*log[6*i + √(-35)] = -i*log(11.916*i) = 1.57 - 2.48*i

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      First, take the inverse sine of both sides of the equation. That gives you x = sin-1(6), which is sadly undefined...in reality, but who needs that! It can be proven that sin-1(x) = -i*log[i*x + √(1-x2)] So in this case: = -i*log[i*6 + √(1-36)] = -i*log[6*i + √(-35)] = -i*log(11.916*i) = 1.57 - 2.48*i

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