# Trigonometric Equation

## Homework Statement

cos(x)+2tg(x) = 7/(4*cosx(x)) Solve for x on the real numbers set.

## Homework Equations

tg(x) = sin(x)/cos(x) ,
cos^2(x) = 1-2*sin^2(x) ,
The determinant is D = b^2 - 4ac ,
Also cos(x) ,tg(x), sin(x) | -1 < x < 1

## The Attempt at a Solution

For final arrangement I got is:
-8sin^2(x) + 8sin(x) - 3 = 0

But the determinant is 8^2 - 4*(-8)*(-3) which is 64 - 96 thus the equation doesn't have a real solution.

I checked on WolframAlpha and it has some weird solutions and I was also told that it does have a real solution.

ehild
Homework Helper
1.

## The Attempt at a Solution

For final arrangement I got is:
-8sin^2(x) + 8sin(x) - 3 = 0

The coefficient 8 in front of sin^2(x) is wrong.

ehild

Why?
Let's say we rearrange the original equation to: cos^2(x) + 2 sin(x) = 7/4
// We multiplied by cosx(x) //
Then cos^2(x) becomes 1-2sin^2(x)... That multiplied by 4 will be 4-8sin^2(x).
I don't think where I got that wrong.

tiny-tim
Homework Helper
Hi Icelove!

(try using the X2 tag just above the Reply box )
cos^2(x) = 1-2*sin^2(x)

nooo … learn your trigonometric identities …

cos2x = 1 - 2sin2x

ehild
Homework Helper
$$\cos{x}\cdot\cos{x}=(\cos{x})^2=1-(\sin{x})^2$$

ehild

cos(x)+2tg(x) = 7/(4*cosx(x)) Solve for x on the real numbers set.

By this you the complete equation is

$$(4 \cdot cos(x)^2)\cdot cos(x) + (4\cdot cos(x)^2)\cdot (2tan(x)) - 7 = 0$$ which implies that

$$(4 \cdot cos(x)^2) \cdot (cos(x) + 2 tan(x)) - 7 = 0$$

Last edited:
ooooooooooooooooohh... Thanks. :D