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Answer 1 · 2018-01-22T14:05:42

# x =(-1)^k(-pi/2)+kpi=kpi+(-1)^(k+1)(pi/2), k in ZZ#.

#sin^2x+sinx-2=0#.

#:. ul(sin^2x+2sinx)-ul(sinx-2)=0#.

#:. sinx(sinx+2)-1(sinx+2)=0#.

#:. (sinx+2)(sinx-1)=0#.

#:. sinx=-2,# which is not possible for any #x in RR,# or,

#sinx=-1=sin(-pi/2)#.

Now, #sintheta=sinalpha rArr theta=(-1)^kalpha+kpi, k in ZZ#,

#:. x =(-1)^k(-pi/2)+kpi, k in ZZ, or, #

#x=kpi+(-1)^(k+1)(pi/2), k in ZZ#.

Answer 2 · 2018-01-22T14:12:53

#x=90^@# or #pi/2 " radians"#.

Take #sinx=y#.

Now, we have #y^2+y-2=0#

Here, #a=1,b=1,c=-2#

Use the quadratic formula #(-b+-sqrt(b^2-4ac))/(2a)#.

#(-1+-sqrt(1+8))/(2*1)#

#(-1+-3)/2#

#y=(-1-3)/2,(-1+3)/2#

#y=-2,1#

However, #sinx# can never be negative. So, #sinx=1#

#sinx=1#

#x=sin^-1 1#

#x=90^@# or #pi/2 " radians"#.