Calculus and Trigonometric Identities

Question

Find all the values of x in the interval [0, 2π] that satisfy the equation sin(4x)=cos(2x).

Answer 1

sin(4x) = cos(2x)

Recall that sin(2θ) = 2sinθcosθ

So, 2sin(2x)cos(2x) = cos(2x)

2sin(2x)cos(2x) - cos(2x) = 0

cos(2x) [2sin(2x) - 1] = 0

So, cos(2x) = 0 or sin(2x) = 1/2

If cos(2x) = 0, then 2x = π/2 + kπ. So, x = π/4 + kπ/2, k = 0, 1, -1, 2, -2, ...

Plug in k = 0,1,2,and 3 to get x = π/4, 3π/4, 5π/4, 7π/4

If sin(2x) = 1/2, then 2x = π/6 + 2kπ or 5π/6 + 2kπ. So, x = π/12 + kπ or 5π/12 + kπ, k = 0, 1, -1, 2, -2, ...

Plug in k = 0,1 to get x = π/12, 5π/12, 13π/12, 17π/12

So, there are a total of 8 solutions in the interval [0, 2π]:

π/4, 3π/4, 5π/4, 7π/4, π/12, 5π/12, 13π/12, 17π/12

Answer 2

sin(4x)=cos(2x) Double angle formula: sin(2x)=2sinxcosx

2sin(2x)cos(2x)=cos(2x)

2sin(2x)cos(2x)-cos(2x)=0

cos(2x)(2sin(2x)-1)=0

cos(2x)=0 , 2sin(2x)-1=0

2x=π/2, 2x=3π/2 2sin(2x)-1=0

x=π/4, x=3π/4 sin(2x)=1/2

2x=π/6, 2x=5π/6

x=π/12, x=5π/12

All the values of x in the interval [0,2π] are : x=π/12,5π/12,π/4,3π/4

Answer 3

sin 4x = cos 2x

sin 2(2x) = cos 2x

2 sin 2x cos 2x = cos 2x

Can you complete and answer?