Given that 5cos^2(x) - cos(x) = sin^2(x), find the possible values of cos(x) using a suitable quadratic equation.

First, need to get all the terms in the equation to be the same. Using the following identity, it is possible to achieve this:

sin2(x) + cos2(x) = 1

1 - cos2(x) = sin2(x)

Substituting this into the equation in the question:

5cos2(x) - cos(x) = 1 - cos2(x)

6cos2(x) - cos(x) - 1 = 0

Replace the term cos(x) with y:

6y2 - y - 1 = 0

Product = -6

Sum = -1

There numbers that satisfy this are -3 and 2. Therefore, the factorised form of the eqation is:

(2y - 1)(3y + 1) = 0

The roots of this equation are: y = cos(x) = -1/3 or 1/2. Therefore these are the possible values of cos(x).

Answered by Andrew B. Maths tutor

6321 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Find the stationary points of y= 5x^2 + 2x + 7


If 1/(x(a-x)) is equivalent to B(1/x + 1/(a-x)), Express 'B' in terms of 'a'


Given y = ln((2x+3)/(7x^3 +1)). Find dy/dx


(x-4)^3


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 2013–2024

Terms & Conditions|Privacy Policy