Solve the equation for exact solutions over the interval [0,360) for 9sin^2x-6sinx=1 ?

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Answer 1 · 2018-02-09T20:23:16

#color(blue)(x= 53.58^@ , 126.42^@ , 187.94^@, 352.06^@#

#9sin^2x-6sinx=1#

#9sin^2x-6sinx-1=0#

Let #color(white)(88)u=sinx#

Then:

#9u^2-6u-1=0#

Using Quadratic Formula:

#u=(-(-6)+-sqrt((-6)^2-(4*9*-1)))/(2*9)#

#u=(6+-sqrt(36+36))/18=(6+-6sqrt(2))/18=(1+-sqrt(2))/3#

#u=sinx=(1+sqrt(2))/3=>x=arcsin((1+sqrt(2))/3)=53.58^@ , 126.42^@#

And:

#x=arcsin((1-sqrt(2))/3)=187.94^@ , 352.06^@#

Solutions:

#color(blue)(x= 53.58^@ , 126.42^@ , 187.94^@, 352.06^@#