# Trigonometric equation solving 2cos x=tan x

• AN630078
Thank you for catching that. In summary, the conversation discusses using Desmos to plot a graph and finding the intersection points of a trigonometric equation. The conversation also covers using trigonometric identities and the quadratic formula to solve for the solutions in a given range. There is also mention of using a calculator and a clarification on a typo.

#### AN630078

Homework Statement
I have really been rather struggling with trigonometric equations, hwoever I have been teaching myself and practising to improve and hone my understanding. Although, I remain a little uncertain in areas. I have attempted some questions below but was wondering if anyone could offer me some advice on how to improve my workings or apply more suitable methods.

Sketch the graphs y=2cos x and y=tan x.
a. How many values of x satisfy the equation in the range 0 ≤θ ≤360 degrees?
b. Show that the x-values at the points of intersection satisfy the equation 2sin^2x+sins-2=0
c. Solve the equation 2cos x=tan x between 0 and 360 degrees
Relevant Equations
tan x=sin x / cos x
sin^2x+cos^2x=1
a. I have just plotted the graph using desmos and attached an image here. Clearly, there are two values of x that satisfy the equation in the range. Do I need to add anything to this statement, I feel the response is a little brief for the question?

b. Using the trigonometric identities;
tan x=sin x / cos x
and sin^2x+cos^2x=1

2cos x=tan x
Multiply the whole equation by cos x:
2cos x=sinx
Using the identity sin^2x+cos^2x=1, 2cos x becomes: 2(1-sin^2x)
2(1-sin^2x)=sinx
Expand the brackets;
2-2sin^2x=sinx
Subtract sin x from both sides;
2-2sin^2x-sinx=0
Divide by -1;
2sin^2x+sinx-2=0

c. To solve the equation 2cos x=tan x
This is shown to be equal to the quadratic;
2sin^2x+sinx-2=0

Let sin x = u
2u^2+u-2=0

Using the quadratic formula; a=2, b=1, c=-2
u=-b±√b^2-4ac/2a
u=-1±√1^2-4*2*(-2)/2*2
u=-1±√17/4
u=0.780 and x=-1.28 to 3.s.f

sin x = -1±√17/4
x=arcsin 0.78 =51.3 degrees to 3.s.f
x= arcsin -1.28 = no real solutions as x cannot be smaller than 1 for real solutions

To find other solutions in the range 0 ≤θ ≤360 degrees use sin θ=sin(180-θ):
180-51.3=128.7 degrees

So the solutions are 51.3 and 129 degrees to 3.s.f?

Would this be correct. I am very uncertain of trigonometric equation problems and I have been teaching myself which is perhaps why my knowledge is a little unstable and I am am sure in places erroneous. I would be very grateful of any help

#### Attachments

• Question 2 a.png
36.8 KB · Views: 118
With respect to (a) shouldn't your graph use degrees in the x-axis as it asks for the intersection points with the domain of 0 to 360 degrees and your plot doesn't show that.

For (b), you start with ##2 cos(x) = tan(x)## and then multiply both sides by ##cos(x)## and you should get ##2 cos^2(x) = sin(x)## right?

AN630078
AN630078 said:
So the solutions are 51.3 and 129 degrees to 3.s.f?

Would this be correct.
Have you got a calculator?

AN630078 said:
b. Using the trigonometric identities;
tan x=sin x / cos x
and sin^2x+cos^2x=1

2cos x=tan x
Multiply the whole equation by cos x:
2cos x=sinx
You have a typo above. It should be ##2\cos^2(x) = \sin(x)##. I think you just neglected to add the exponent here, because in subsequent work you have the exponent.

Last edited:
AN630078
jedishrfu said:
With respect to (a) shouldn't your graph use degrees in the x-axis as it asks for the intersection points with the domain of 0 to 360 degrees and your plot doesn't show that.

For (b), you start with ##2 cos(x) = tan(x)## and then multiply both sides by ##cos(x)## and you should get ##2 cos^2(x) = sin(x)## right?
Thank you for your reply. Yes you are correct, I just quickly jotted down the graph from Desmos, in actually sketching it I would use degrees on the x-axis.

Sorry, for (b) that is a typo, I meant ##2cos^2(x)=sin(x)## .

Last edited by a moderator:
PeroK said:
Have you got a calculator?
Yes, why?

Mark44 said:
You have a typo above. It should be ##2\cos^2(x) = \sin(x)##. I think you just neglected to add the exponent here, because in subsequent work you have the exponent.
Yes you are correct, it was a typo, I meant ##2cos^2(x)=sin(x)##

Last edited by a moderator: