How do I solve the equation "2cos(x) = sin(2x), for 0 ≤ x ≤ 3π"?

  • Google+ icon
  • LinkedIn icon
  • 3090 views

The key to solving this equation is realizing that sin(2x) can be written in terms of sin(x) and cos(x) using a double-angle formula. (With trigonometric problems similar to this one, you should always check if any trigonometric identites like the double-angle formulae can be used, as these can often help you.)

Using your IB formula booklet, you will see that the double-angle formula for sine is:

sin(2x) = 2sin(x)cos(x) 

Therefore, we can rewrite the given equation from:

2cos(x) = sin(2x)

to:

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

Next, we notice that both sides of the equation are multiplied by 2, so we can divide both sides by 2. This yields the equation:

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

We can now divide both sides of the equation by cos(x), which leaves us with:

sin(x) = 1 

Finally, we must think about the angles at which sin(x) is equal to 1. You should realize, perhaps by imagining the unit circle, that sine is equal to 1 whenever x = π/2 + n2π, where n is any integer. 

However, this is not the final answer, as the problem gave us a restricted domain for x. x must be in between 0 and 3π. Therefore, the only possible values for x are π/2 and 5π/2.

So, the answer is:

x = π/2 and 5π/2

Sanveer R. IB Maths tutor, IB Economics tutor

About the author

is an online IB Maths tutor with MyTutor studying at LSE University

How MyTutor Works

Still stuck? Get one-to-one help from a personally interviewed subject specialist.

95% of our customers rate us

Browse tutors

We use cookies to improve your site experience. By continuing to use this website, we'll assume that you're OK with this. Dismiss

mtw:mercury1:status:ok