Equation on interval

[rachelmaddie]

I need this checked please.


Rewrite the equation;

tan2x (sec2x - 1) + 2sec2x - 2 =0

Lets name: sin(x) = s, cos(x)=c, then

tan(x) = s/c, sec(x) = 1/c, and s2 + c2 =1.

The equation in terms of new variables s and c:
s2/c2 (1/c2 -1) + 2/c2 -2 =0

s2/c2 (1 - c2)/c2 + 2/c2 -2 =0

s4/c4 + 2/c2 -2 =0

s4 + 2c2 -2c4 =0

Replace: s4 = (1- c2)2 =1 - 2c2 + c4

1 - 2c2 + c4 + 2c2 -2c4 =0
1 - c4 =0
c = 1 or c = -1

cos(x) = 1 or cos(x) = -1

Answer: x = 0, x= π

 

[pka]

Look HERE

 

[rachelmaddie]

Look HERE

Is my work correct?

 

[pka]

Is my work correct?

Why do you not decide or yourself?
You have the website solution. They give one more solution than you have.

 

[Dr.Peterson]

Is my work correct?

You're right that correct work is different from having a correct answer, which could be an accident; and pka used the wrong interval, which explains his extra solution.

But before I'll be willing to check your work, you'll have to make an effort to make it readable (in fact, to make it mean what you intend). Please use either the x² button on the toolbar, or "^2", to indicate exponents correctly. (And if you use ^, you can select an expression and click the calculator icon, and it will look better.)

 

[rachelmaddie]

You're right that correct work is different from having a correct answer, which could be an accident; and pka used the wrong interval, which explains his extra solution.

But before I'll be willing to check your work, you'll have to make an effort to make it readable (in fact, to make it mean what you intend). Please use either the x² button on the toolbar, or "^2", to indicate exponents correctly. (And if you use ^, you can select an expression and click the calculator icon, and it will look better.)

Rewrite the equation;

tan^2x (sec^2x - 1) + 2sec^2x - 2 =0

Lets name: sin(x) = s, cos(x)=c, then

tan(x) = s/c, sec(x) = 1/c, and s^2 + c^2 =1.

The equation in terms of new variables s and c:
s^2/c^2 (1/c^2 -1) + 2/c^2 -2 =0

s^2/c^2 (1 - c^2)/c^2 + 2/c^2 -2 =0

s^4/c^4 + 2/c^2 -2 =0

s^4 + 2c^2 -2c^4 =0

Replace: s^4 = (1- c^2)^2 =1 - 2c^2 + c^4

1 - 2c^2 + c4 + 2c^2 -2c^4 =0
1 - c^4 =0
c = 1 or c = -1

cos(x) = 1 or cos(x) = -1

Answer: x = 0, x= π

 

[Dr.Peterson]

I added the simple formatting I mentioned, using the MATH button:

Rewrite the equation;

[MATH]tan^2x (sec^2x - 1) + 2sec^2x - 2 =0[/MATH]
Lets name: [MATH]sin(x) = s[/MATH], [MATH]cos(x)=c[/MATH], then

[MATH]tan(x) = s/c[/MATH], [MATH]sec(x) = 1/c[/MATH], and [MATH]s^2 + c^2 =1[/MATH].

The equation in terms of new variables s and c:
[MATH]s^2/c^2 (1/c^2 -1) + 2/c^2 -2 =0[/MATH]
[MATH]s^2/c^2 (1 - c^2)/c^2 + 2/c^2 -2 =0[/MATH]
[MATH]s^4/c^4 + 2/c^2 -2 =0[/MATH]
[MATH]s^4 + 2c^2 -2c^4 =0[/MATH]
Replace: [MATH]s^4 = (1- c^2)^2 =1 - 2c^2 + c^4[/MATH]
[MATH]1 - 2c^2 + c^4 + 2c^2 -2c^4 =0[/MATH][MATH]1 - c^4 =0[/MATH][MATH]c = 1[/MATH] or [MATH]c = -1[/MATH]
[MATH]cos(x) = 1[/MATH] or [MATH]cos(x) = -1[/MATH]
Answer:[MATH] x = 0, x= π[/MATH]

Looks correct.