# Trigonometry

posted by .

Find all solutions of 4(cos(x))**2-4 sin(x)-5=0 in the interval (6pi, 8pi)

I tried to work it out and got: 4(cos**2x)-4cosx -9 = 0, but I can't figure out what cosx = from there to finish the problem.

• Trigonometry -

________________________________________

Remark :

( cos x ) ^ 2 = 1 - ( sin x ) ^ 2
________________________________________

4 ( cos x ) ^ 2 - 4 sin( x ) - 5 = 0

4 [ 1 - ( sin x ) ^ 2 ] - 4 sin( x ) - 5 = 0

4 - 4 ( sin x ) ^ 2 - 4 sin ( x ) - 5 = 0

- 4 ( sin x ) ^ 2 - 4 sin ( x ) - 1 = 0 Multiply both sides by - 1

4 ( sin x ) ^ 2 + 4 sin ( x ) + 1 = 0
________________________________________

Remark :

( a + b ) ^ 2 = a ^ 2 + 2 a b + b ^ 2

So :

4 sin ( x ) ^ 2 + 4 sin ( x ) + 1 = [ 2 sin ( x ) + 1 ] ^ 2

Becouse :

[ 2 sin ( x ) + 1 ] ^ 2 =

[ 2 sin ( x ) ] ^ 2 + 2 * 2 sin ( x ) + 1 ^ 2 =

4 sin ( x ) ^ 2 + 4 sin ( x ) + 1

4 ( sin x ) ^ 2 + 4 sin ( x ) + 1 = 0 is same :

[ 2 sin ( x ) + 1 ] ^ 2 = 0
________________________________________

[ 2 sin ( x ) + 1 ] ^ 2 = 0 Take the square root of both sides

2 sin ( x ) + 1 = 0 Subtract 1 from both sides

2 sin ( x ) = - 1 Divide bboth sides by 2

sin ( x ) = - 1 / 2

sin ( x ) = - 1 / 2 for :

x = 7 pi / 6 = 210 °

and

x = 11 pi / 6 = 330 °

The period of sin ( x ) is 2 pi

So :

sin ( x ) = - 1 / 2 for :

x = 2 n pi +7 pi / 6

and

x = 2 n pi + 11 pi / 6

n is an integer

In the interval ( 6 pi, 8 pi )

x = 2 * 3 pi + 7 pi / 6

and

x = 2 * 3 pi + 11 pi / 6

Solutions :

x = 6 pi + 7 pi / 6

and

x = 6 pi + 11 pi / 6

OR

x = 6 * 6 pi / 6 + 7 pi / 6 = 36 pi / 6 + 7 pi / 6 = 43 pi / 6

and

x = 6 * 6 pi / 6 + 11 pi / 6 = 36 pi / 6 + 11 pi / 6 = 47 pi / 6

• Trigonometry -

You apparently think that sinx = 1+cosx
The correct identity is cos^2 + sin^2 = 1
so, cos^2 x = 1-sin^2 x

4(1-sin^2 x) - 4sinx - 5 = 0
4 - 4sin^2 x - 4sinx - 5 = 0

4sin^2 x + 4sinx + 1 = 0
(2sinx + 1)^2 = 0
2sinx + 1 = 0
sinx = -1/2

Take it from there.

## Similar Questions

1. ### pre-calc

Solve: cos(2x-180) - sin(x-90)=0 my work: cos2xcos180 + sin2xsin180= sinxcos90 - sin90cosx -cos2x - sin2x= cosx -cos^2x + sin^2x -2sinxcosx=cosx I'm stuck here. I tried subtracting cosx from both sides and making sin^2x into 1- cos^2x, …
2. ### Pre-Calculus

How can this identity be proved/verified?
3. ### Precal

I do not understand how to do this problem ((sin^3 A + cos^3 A)/(sin A + cos A) ) = 1 - sin A cos A note that all the trig terms are closed right after there A's example sin A cos A = sin (A) cos (A) I wrote it out like this 0 = - …
4. ### maths - trigonometry

I've asked about this same question before, and someone gave me the way to finish, which I understand to some extent. I need help figuring out what they did in the second step though. How they got to the third step from the second. …
5. ### pre calc trig check my work please

sin x + cos x -------------- = ? sin x sin x cos x ----- + ----- = sin x sin x cos x/sin x = cot x this is what i got, the problem is we have a match the expression to the equation work sheet and this is not one of the answers. need
6. ### trig

Find all solutions of the equation. Leave answers in trigonometric form. x^5-1024=0 I got 4(cos tehta + i sin tehta), tehta = 0, 2pi/5, 4pi/5, 6pi/5, 8pi/5 is this right
7. ### Trigonometry

Find all solutions of 4 (cos (x)**2)-1=0 in the interval (6pi, 8pi). (Leave your answers in exact form and enter them as a comma-separated list.)