Question

Solve sin4xcos2x-cos4xsin2x=square root of 2 sinx over the interval [0,2pi)

Answer 1

The trigonometric equation is solved using the sine difference identity, simplifying to sin(2x) = √2 sin(x). The solution involves finding values of x that satisfy the equation over the interval [0,2π). Trigonometric identities such as double angle formulas are essential in the process.

Step-by-step explanation:

The student has presented a trigonometric equation involving sine and cosine functions to solve: sin(4x)cos(2x) - cos(4x)sin(2x) = √2 sin(x) over the interval [0,2π). This can be addressed by recognizing the left-hand side as the expansion of the sine difference identity: sin(A - B) = sin(A)cos(B) - cos(A)sin(B), where A = 4x and B = 2x. Therefore, the equation simplifies to sin(2x) = √2 sin(x). We solve this equation over the specified interval by looking for values of x that satisfy the condition.

However, none of the reference equations or principles provided directly align with solving the original question. Hence, we must rely solely on our knowledge of trigonometric identities, such as the double angle formulas which are relevant to this problem. The double angle identities state that sin(2θ) = 2sin(θ)cos(θ) and cos(2θ) = cos^2(θ) - sin^2(θ) or equivalently cos(2θ) = 2cos^2(θ) - 1 and cos(2θ) = 1 - 2sin^2(θ).

Answer 2

`\bf sin({{ \alpha}} - {{ \beta}})=sin({{ \alpha}})cos({{ \beta}})- cos({{ \alpha}})sin({{ \beta}}) \\\\\\ sin(2\theta)=2sin(\theta)cos(\theta)\\\\ -------------------------------\\\\ sin(4x)cos(2x)-cos(4x)sin(2x)=√(2)sin(x) \\\\\\ sin(4x~~-~~2x)=sin(x)√(2)\implies sin(2x)=sin(x)√(2) \\\\\\`


`\bf 2sin(x)cos(x)=sin(x)√(2)\implies 2sin(x)cos(x)-sin(x)√(2)=0 \\\\\\ sin(x)~[2cos(x)-√(2)]=0\\\\ -------------------------------\\\\ sin(x)=0\implies \measuredangle x=0~~,~~\pi \\\\ -------------------------------\\\\ 2cos(x)-√(2)=0\implies 2cos(x)=√(2)\implies cos(x)=\cfrac{√(2)}{2} \\\\\\ \measuredangle x=(\pi )/(4)~~,~~(7\pi )/(4)`

now, we're not including the III and II quadrants, where the cosine has an angle of the same value, but is negative, because the exercise seems to be excluding the negative values of √(2).