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Trig. equations help needed?

+1 vote

I need to find all solutions of 4(cos(x))^2 -4 = 0 in interval [4pi,6pi]

I also need to find all solutions of 4(sin(x))^2 - 2sin(x) - 2 = 0 in interval [4pi, 6pi]

Please leave these in exact form, that would help a lot!!!!!!

asked Jan 17, 2013 in TRIGONOMETRY by angel12 Scholar

1 Answer

+2 votes

4 cos2x - 4 = 0

Divide each side by 4.

[ 4 cos2x - 4 ] / 4 = 0 / 4

(4 cos2x) / 4 - 4 / 4 = 0

cos2x - 1 = 0

Add 1 to each side

cos2x - 1 + 1 = 0 + 1

cos2x  = 1

Apply square root each side

√( cos2x)  = √1

Note : √A = ±A and √A2 = A

So,

cos x = ± 1

cos x = 1  or cos x = -1

In the interval [0,2π],

When cosx = 1 then x = 0 and 2π,

So in the interval [4π,6π]

x = 4π and 6π.

Similarly,

When cosx = -1 in the 1st interval then x = π

So in the later interval, x = 4π + π = 5π

 

4 cos2x - 4 interval [4π, 6π]

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

4 sin2x - 2sinx - 2 = 0

Let sin x = t

Then

4t2 - 2t - 2 = 0

Now solve the equation factor method

4t2 - 4t + 2t - 2 = 0

4t(t - 1) + 2 (t - 1) = 0

Take out common factors.

(4t + 2)(t - 1) = 0

4t + 2 = 0  or t - 1 = 0

4t + 2 = 0

Subtract 2 from each side.

4t + 2 - 2 = 0 - 2

4t = - 2

Divide each side by 4

4t / 4 = - 2 / 4

t = - 1 / 2

And

t - 1 = 0

Add 1 to each side

t - 1 + 1 = 1

t = 1

But t = sin x

So, sin x = - 1 / 2 and sin x = 1

As the sine (and cosine) function has a cycle of 2π radians

Then sinx = -1/2 in the interval [0, 2π]

When x = 7π/6 and 11π/6 therefore in the interval [4π, 6π],

x = (4π + 7π/6)

= 31π/6

And (4π + 11π/6)

= 35π/6.

In the interval [0, 2π], when sin x = 1

Then x = π/2 so in the interval [4π, 6π],

x = 4π + π/2

= 9π/2.

answered Jan 22, 2013 by richardson Scholar
you really good at trigonometry:)

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