Solve the equation in the interval [0,360) for: ?

Question

#11sin^2x-7sinx=2#

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Answer 1 · 2018-02-11T11:16:37

#x~~58.23,121.76,192.35,347.65#

We want to solve for x in the interval #[0,360)# (degrees)

#11sin^2(x)-7sin(x)-2=0#

Let #u=sin(x)# and solve as a traditional quadratic equation

#11u^2-7u-2=0#

By the quadratic formula

#u=(7+-sqrt((-7)^2-4*11(-2)))/(2*11)#

#=(7+-sqrt(137))/22#

Substitute #u=sin(x)#

#sin(x)=(7+-sqrt(137))/22#

By the inverse sine

#x~~58.23^@,121.76^@,192.35^@,347.65^@#