Solve the equation tanx/cosx = 1 for 0°<x<360°

Firstly we need to rearrange this equation so that it contains only one trigonometric function of x (i.e. tanx, cosx or sinx) which will make it much easier to solve.

We can do this in the following way:

First multiply both sides by cosx in order to get tanx = cosx.
It then helps to write tanx in terms of cosx and sinx (tanx = sinx/cosx) and if we put this into the equation we now have, we get sinx/cosx = cosx.
Then multiply both sides by cosx a second time, to get sinx=cos2x.
Now we know from rearranging the identity sin2x + cos2x = 1, that cos2x=1-sin2x, and so if we substitute this into our equation we get sinx=1-sin2 x.
This gives us sin2x+sinx-1=0, which gives an equation recognisable as a quadratic equation, all in terms of sinx, meaning it will now be far easier to solve.

To solve:

Trigonometric quadratic equations can sometimes look much more complicated to solve that a quadratic in the form ax2+bx+c=0 but they are really no different and just take some getting used to. In this case, instead of x coefficients, we have sinx coefficients, meaning our equation is in the form a(sinx)2+b(sinx)+c=0.

Like any other quadratic equation we can use the quadratic formula where a=1, b=1 and c=-1, which gives us the solutions sinx=(-1+sqrt5)/2 and sinx=(1+sqrt5)/2.

We then have to go back to the question and see that we are looking for solutions where x is between 0° and 360°. The clearest way to see where our solutions are is to draw the graph y=sinx with x axis from 0° to 360° and see for which values of x, y=(-1+sqrt5)/2 and for which values y=(1+sqrt5)/2.  We can immediately see that (1+sqrt5)/2 > 1 so there is no solution to sinx=(1+sqrt5)/2 because sinx is bounded above by 1 (i.e. can’t any value higher than 1)

To get the specific values for x where sinx=(-1+squrt5)/2, we can use arctan in the calculator, and then check from the graph we have drawn whether there are any other solutions in the domain.

If we do this, we see that x=38.2° and x=142° are the solutions to sinx=(-1+sqrt5)/2  in the domain.

So our solutions are x=38.2° and x=142° (rounded to 3sig.figures)

Answered by Linetta A. Maths tutor

22063 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

((x^2+4x)/2x)-((x^2-4x)/x)


Using the addition formula for sin(x+y), find sin(3x) in terms of sin(x) and hence show that sin(10) is a root of the equation 8x^3 - 6x + 1


Solve for 0=<x<360 : 2((tanx)^2) + ((secx)^2) = 1


What is a moment and how do I calculate it?


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