Solve for x sin(4x)- square root of 2sin(2x)=0

Problem

sin(4*x)−√(,2)*sin(2*x)=0

Solution

1. Apply the double angle identity for sine, sin(2*θ)=2*sin(θ)*cos(θ) to the term sin(4*x) by treating 4*x as 2*(2*x)

sin(4*x)=2*sin(2*x)*cos(2*x)

2. Substitute this identity back into the original equation.

2*sin(2*x)*cos(2*x)−√(,2)*sin(2*x)=0

3. Factor out the common term sin(2*x) from the expression.

sin(2*x)*(2*cos(2*x)−√(,2))=0

4. Set each factor to zero to find the possible solutions for x

sin(2*x)=0

2*cos(2*x)−√(,2)=0

5. Solve the first equation sin(2*x)=0 The sine function is zero at integer multiples of π

2*x=n*π

x=(n*π)/2

6. Solve the second equation 2*cos(2*x)−√(,2)=0 by isolating the cosine term.

cos(2*x)=√(,2)/2

7. Find the general solution for 2*x where the cosine is √(,2)/2 which occurs at ±π/4+2*n*π

2*x=±π/4+2*n*π

x=±π/8+n*π

Final Answer

sin(4*x)−√(,2)*sin(2*x)=0⇒x=(n*π)/2,x=π/8+n*π,x=−π/8+n*π

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