Question #9b19b

1 Answer
Jan 7, 2018

#x=sin^-1(sqrt5 /5 )#

Explanation:

Assuming your problem is: #4cos^2x=sin^2x+3#

Using #sin^2x+cos^2x=1#, we get

#4(1-sin^2x)=sin^2x+3#

#4-4sin^2x=sin^2x+3#

#5sin^2x=1#

#sin^2x=0.2#

#=> sinx = sqrt0.2 = sqrt5 / 5 #

#x=color(red)(sin^-1(sqrt5 / 5 )~~0.4647# # color(red)(rad#