How do I solve the equation for solutions on the interval of 0 < x < 2𝜋 for the equation 2sin(x) cos(x) - sin (x) = 0?

Question

I am really confused on how to work out this equation.

Answer 1

Good Analysis, but. x = 2π would also work with sin(x) = 0.

However, both x = 0 and 2π are out because of the domain given: 0<x<2π

Answer 2

The big trick here is to recognize that you need to factor out the sin(x). Recognizing this comes with time and practice.

So if we factor out sin(x) we get sin(x)(2cos(x) - 1)=0.

Notice how we have two parentheses here. If you plug in a number for x, sin(x) will be just one number, and 2cos(x)-1 will also just be one number. So both parentheses could be thought of as one number.

So we have two numbers that multiply to get 0. You might know that if two numbers multiply to 0, one of them must be zero, so we set each parentheses equal to 0.

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

The right equation can be turned into cos(x) = 1/2 through adding 1 to both sides, and dividing both sides by 2. So now we have:

sin(x) = 0 and cos(x) = 1/2

There are many ways to find x from here. I like to find them manually using a thing called reference triangles (though it's too long to explain here), but I find the most common method is to use a table, memorize the unit circle, or use a calculator. However you want to find it, sin(x)=0 happens when x = 0 and x = π, and cos(x) = 1/2 happens when x = 60 degrees and x = 300 degrees. (Which are π/3 and 5π/3, respectively, in radians)

Let me know if this helps! :)

Answer 3

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

factor sin(x)

Now, solve these separately...

sin(x) = 0 → x = 0, π

2cos(x) - 1 = 0 → 2cos(x) = 1 → cos(x) = 1/2 → x = π/3, 5π/3

+1 ÷2

Thus, the values on interval 0 < x < 2π that make 2sin(x)cos(x) - sin(x) = 0 are

x = 0, π/3, π, 5π/3