The number of real solution of the given equation #color(blue)(cos^(7)x + sin^(4)x=1)# is 3 in the interval #[-pi,pi]# as revealed from the graph.
Solutions are: #color(red)({-pi/2,0and pi/2})#
Trying to solve now
#color(blue)(cos^(7)x + sin^(4)x=1)#
#=>cos^(7)x -1+ sin^(4)x=0#
#=>cos^(7)x -(1 sin^(4)x)=0#
#=>cos^(7)x -(1 sin^2x)(1+sin^2x)=0#
#=>cos^(7)x -cos^2x(1+sin^2x)=0#
#=>cos^(2)x(cos^5x -1-sin^2x)=0#
#=>cos^(2)x(cos^5x -1-1+cos^2x)=0#
#=>cos^(2)x(cos^5x -2+cos^2x)=0#
So #cos^2x=0#
#=>cosx=0#
This gives two solutions
#x=-pi/2,pi/2# in the interval #[-pi,pi]#
Again
#=>(cos^5x +cos^2x-2)=0#
#=>(cos^5x-cos^4x +cos^4x-cos^3x+cos^3x-cos^2x+2cos^2x-2cosx+2cosx-2)=0#
#=>cos^4x(cosx-1)+cos^3x(cosx-1) +cos^2x(cosx-1)+2 cosx(cosx-1)+2(cosx-1)=0#
#=>(cosx-1)(cos^4x+cos^3x+cos^2x+2 cosx+2)=0#
So #cosx-1=0#
#=>cosx=1#
This gives a solution #x=0# in the interval #[-pi,pi]#
