How do you verify that the x values pi/16, (3pi)/16π16,3π16 are solutions to 2cos^2 4x-1=02cos24x−1=0?
1 Answer
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)^2Note that cos24x=(cos4x)2
color(blue)(x=pi/(16))x=π16
"left side " =2cos^2(4xxpi/(16))-1left side =2cos2(4×π16)−1
=2(cos(pi/4))^2-1=2xx(1/sqrt2)^2-1=2(cos(π4))2−1=2×(1√2)2−1
=2xx1/2-1=0=" right side, hence a solution"=2×12−1=0= right side, hence a solution
color(blue)(x=(3pi)/16)x=3π16
"left side " =2cos^2(4xx(3pi)/16)-1left side =2cos2(4×3π16)−1
=2(cos((3pi)/4))^2-1=2(cos(3π4))2−1
=2(-cos(pi/4))^2-1=2(−cos(π4))2−1
=2xx(-1/sqrt2)^2-1=2×(−1√2)2−1
=2xx1/2-1=0=" right side, hence a solution"=2×12−1=0= right side, hence a solution
