Question #70e43

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Answer 1 · 2018-01-24T10:01:14

#x=90°, x=210°, x=330°#

We want find the solution to:

#2*sin^2(x)-sin(x)-1=0#

Let #u=sin(x)#

#2*u^2-u-1=0#

Solve the quadratic equation

#u=(-(-1)+-sqrt((-1)^2-(4*2*(-1))))/(2*2)=(1+-3)/4#

Substitute #u=sin(x)#

#sin(x)=(1+-3)/4#

Therefore

#sin(x)=-1/2# or #sin(x)=1#

Find the values for x in the interval

#x=90°, x=210°, x=330°#