Question #b1e4e

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Answer 1 · 2018-01-16T11:55:13

See below.

Identity.

#color(red)(sin(2x)=2sin(x)cos(x))#

#sin^2(2x)-2sin(x)cos(x)-12=0#

Substituting:

#sin^2(2x)-sin(2x)-12=0#

Let #u=sin(2x)#

Then:

#u^2-u-12=0#

#(u-4)(u+3)=0=>u=4 and u=-3#

#u=sin(2x)#

#sin(2x)=4# , #sin(2x)=-3#

The range of #sin(x)# is #color(white)(888)-1 <=x <=1#

This equation has no real solutions.

The graph verifies this.

enter image source here