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- 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

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