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How do you solve 2 sin2-5 sin plus 2 equals 0? - Answers

2*sin^2(x) - 5*sin(x) + 2 = 0 is a quadratic equation in sin(x).therefore,{2*sin(x) - 1}*{sin(x) - 2)} = 0=> sin(x) = 1/2 or sin(x) = 2The second solution is rejected since sin(x) cannot exceed 1.The principal solution is x = arcsin(1/2) = pi/6 radians. Additional or alternative solutions will depend on the domain for x - which has not been given.



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How do you solve 2 sin2-5 sin plus 2 equals 0? - Answers

https://math.answers.com/math-and-arithmetic/How_do_you_solve_2_sin2-5_sin_plus_2_equals_0

2*sin^2(x) - 5*sin(x) + 2 = 0 is a quadratic equation in sin(x).therefore,{2*sin(x) - 1}*{sin(x) - 2)} = 0=> sin(x) = 1/2 or sin(x) = 2The second solution is rejected since sin(x) cannot exceed 1.The principal solution is x = arcsin(1/2) = pi/6 radians. Additional or alternative solutions will depend on the domain for x - which has not been given.



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https://math.answers.com/math-and-arithmetic/How_do_you_solve_2_sin2-5_sin_plus_2_equals_0

How do you solve 2 sin2-5 sin plus 2 equals 0? - Answers

2*sin^2(x) - 5*sin(x) + 2 = 0 is a quadratic equation in sin(x).therefore,{2*sin(x) - 1}*{sin(x) - 2)} = 0=> sin(x) = 1/2 or sin(x) = 2The second solution is rejected since sin(x) cannot exceed 1.The principal solution is x = arcsin(1/2) = pi/6 radians. Additional or alternative solutions will depend on the domain for x - which has not been given.

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      2*sin^2(x) - 5*sin(x) + 2 = 0 is a quadratic equation in sin(x).therefore,{2*sin(x) - 1}*{sin(x) - 2)} = 0=> sin(x) = 1/2 or sin(x) = 2The second solution is rejected since sin(x) cannot exceed 1.The principal solution is x = arcsin(1/2) = pi/6 radians. Additional or alternative solutions will depend on the domain for x - which has not been given.

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