How do you verify that the x values #pi/16, (3pi)/16# are solutions to #2cos^2 4x-1=0#?
1 Answer
Jan 12, 2017
given values are solutions to the equation.
Explanation:
Substitute each of the values into the left side of the equation and if they are solutions then the left side should equal zero, the right side of the equation.
#"Note that " cos^2 4x=(cos4x)^2#
#color(blue)(x=pi/(16))#
#"left side " =2cos^2(4xxpi/(16))-1#
#=2(cos(pi/4))^2-1=2xx(1/sqrt2)^2-1#
#=2xx1/2-1=0=" right side, hence a solution"#
#color(blue)(x=(3pi)/16)#
#"left side " =2cos^2(4xx(3pi)/16)-1#
#=2(cos((3pi)/4))^2-1#
#=2(-cos(pi/4))^2-1#
#=2xx(-1/sqrt2)^2-1#
#=2xx1/2-1=0=" right side, hence a solution"#
