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).

AB

Related Maths A Level answers

All answers ▸

A curve has the equation y=3 + x^2 -2x^3. Find the two stationary points of this curve.


How was the quadratic formula obtained.


How do I find the stationary points of a curve?


Using the parametric equations x=6*4^t-2 and y=3*(4^(-t))-2, Find the Cartesian equation of the curve in the form xy+ax+by=c