Answer:
A
Step-by-step explanation:
We want to solve the equation:
[tex]5\sin(2x)=3\cos(x)[/tex]
To do so, we can rewrite the equation.
Recall the double-angle identity for sine:
[tex]\sin(2x)=2\sin(x)\cos(x)[/tex]
By substitution:
[tex]5\left(2\sin(x)\cos(x)\right)=3\cos(x)[/tex]
Distribute:
[tex]10\sin(x)\cos(x)=3\cos(x)[/tex]
We can subtract 3cos(x) from both sides:
[tex]10\sin(x)\cos(x)-3\cos(x)=0[/tex]
And factor:
[tex]\cos(x)\left(10\sin(x)-3\right)=0[/tex]
Hence, our answer is A.
*It is important to note that we should not divide both sides by cos(x) to acquire 10sin(x) = 3. This is because we need to find the values of x, and one or more may result in cos(x) = 0 and we cannot divide by 0. Hence, we should subtract and then factor.
