How do you solve #sin^2x-4sinx-5=0# and find all general solutions?

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Answer 1 · 2016-07-12T14:38:03

#x = (3pi)/2 + 2n pi# #{n in Z}#

#sin^2x-4sinx-5 =0#

Let #phi -=sin x -># #phi^2 -4phi -5 =0#

Factorise: #(phi+1)(phi-5) = 0#

Therefore #phi =# either #-1 or 5#
Since #phi = sinx -># #-1<= phi <= 1# Therefore #phi# cannot #= 5#

Hence #phi =-1#

Since #phi = sin x#

#sin x =-1#
Therefore #x = arcsin(-1) = (3pi)/2# for #x in (0,2pi)#

Since the question asked for a genral solution:
#x = (3pi)/2 +2n pi# #{n in Z}#
Since the period of the #sin# function is #2pi#.

Answer 2 · 2016-09-04T07:09:02

#x=(3pi)/(2)#

We have: #sin^(2)(x)-4sin(x)-5=0#

This trigonometric equation is in the form of a quadratic, so let's factorise it:

#=> sin^(2)(x)+sin(x)-5sin(x)-5=0#

#=> sin(x)(sin(x)+1)-5(sin(x)+1)=0#

#=> (sin(x)-5))(sin(x)+1)=0#

#=> sin(x)=5 =># undefined

or

#=> sin(x)=-1 => x=(3pi)/(2)#

Therefore, the solution to the equation is #x=(3pi)/(2)#.