Solve sin (2x)

Question

Solve: sin(2x) = √2 cos x

 

0 ≤ x ≤ π

 

Thank you for your help!

Answer 1

Solving this problem requires the very useful "double-angle formulas." The double-angle formula for sin(2x) looks like this.

 

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

 

Substitute this identity in the given equation.

 

sin(2x) = √2cos(x)

2sin(x)cos(x) = √2cos(x)

 

Notice that there's cos(x) on both sides, so I can divide both sides by cos(x). Hold on though, because it's usually not safe to divide by variable expressions. If you divide by variables, you could potentially be dividing by zero, and no one should ever divide by zero.

 

Therefore, we first need to see what value of x makes cos(x) zero, before we can divide by cos(x). That value of x is π/2. This means one of your solutions is x = π/2. This is the only value of x in the interval 0 < x < π, that makes cosx equal to zero.

 

Now we can solve for the second solution. For this scenario, we don't know what x is, but we do know it has a different solution, and therefore for this case, cos(x) does not equal zero anymore, so now it's safe to divide by cosx

 

2sin(x)cos(x) = √2cos(x)

2sin(x) = √2

sin(x) = √2/2

 

This sine value is positive, and sine values are positive in both quadrant I and quadrant II. In order to have sine values with a positive √2/2, this involves a right triangle with a reference angle of 45 degrees in both those quadrants. To get a triangle with a 45 degree reference angle in quadrant I, x = π/4, which is 45 degrees. To get a 45-45-90 triangle in quadrant II, then x has to be 135 degrees, which is 3π/4 in radians.

 

To sum things up, x = π/2, π/4, and 3π/4. I hope this helps, and good luck with your trigonometry adventures!