Answers>Maths>IB>Article

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

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

Answered by Sanveer R. Maths tutor

32246 Views

See similar Maths IB tutors

Related Maths IB answers

All answers ▸

dy/dx = 10exp(2x) - 4; when x = 0, y = 6. Find the value of y when x = 2.


In the arthmetic sequence, the first term is 3 and the fourth term is 12. Find the common difference (d) and the sum of the first 10 terms.


How do you integrate xln(x) between the limits of 0 and 2?


How to find the derivative of sqrt(x) from first principles?


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences