How do you solve? #(4/3)^cosx=sinx#

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Answer 1 · 2017-10-10T19:33:27

See below.

If #x_0# is a solution then

#sin(x_0) gt 0 rArr x_0 in (0,pi)#
#0 lt (4/3)^cos (x_0) le 1 rArr x_0 in [pi/2,pi)# and then

#x_0 in [pi/2,pi)#

and one solution is for #x_0 = pi/2# because

#(4/3)^cos(pi/2) = sin(pi/2)=1#

The other solution in #(pi/2,pi)# must be obtained using an iterative method, giving

#x_1 approx 2.0943951023931953#