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

Question

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.

Answer 1

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Remark :

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

Answer 2

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.