solve 4cos^2x=3 in the interval [0,2pi); give exact answers?

Question

can somebody please go through the steps of solving this problem? i'm having trouble understanding how to work it out.

Answer 1

Just as some of the other answerers have said,
from 4cos^2(x) = 3, you get cos^2(x) = 3 / 4 and
so cos(x) = ± sqrt(3) / 2.

I'm imagining that this is where your trouble starts.
Once you have the basics, it's useful to remember
numbers such as sqrt(3) / 2, sqrt(2) / 2 and 1 / 2,
when it comes to working out angles, as they come
in handy lots of times.

Draw a right-angled triangle with legs equal to 1 unit,
then by Pythagoras, the hypotenuse = sqrt(2) units.
This triangle is half of a unit square, so the 2 equal
angles in the triangle must each be 45º.
Knowing that sin = opp / hyp and cos = adj / hyp,
we have : sin 45º = 1 / sqrt(2) = sqrt(2) / 2, and
this is also true for cos 45º. Also, tan 45º = 1.

Now draw an equilateral triangle with each side equal
to 2 units. Then halve it by constructing a line from
one vertex to the mid-point of the opposite side.
This presents us with 2 right-angled triangles.
Using Pythagoras again on either of the 2 triangles
formed, we find that the hypotenuse = 2 units, while
one leg = 1 unit and the other leg = sqrt(3) units.
The original equilateral triangle has all angles equal
to 60º, so each half-triangle will have the angles 90º,
60º and 30º.
Now we have sin 30º = cos 60º = 1 / 2 and
sin 60º = cos 30º = sqrt(3) / 2.
Also, tan 30º = sqrt(3) / 3 and tan 60º = sqrt(3).

These commonly used angles have relatively simple
ratios. Other angles have more complex exact values.

Back to your question, we have cos(x) = ± sqrt(3) / 2,
or what is the same, ± cos(x) = sqrt(3) / 2.
As we have seen, the principal value of x is 30º. The
principal value is the least positive angle for which the
equation is true.

To find more angles, there are a couple of relations you
probably know. One is that cos(A) = cos(-A). This says
that cos(30º) = cos(-30º) = cos(360º - 30º) = cos(330º).
Here, 30º is in Quadrant I and 330º is in Quadrant 4.

The other relation is : -cos(A) = cos(180º ± A).
Thus, -cos(30º) = cos(180º ± 30º) = cos(150º) or cos(210º).
As we've seen, it doesn't matter if cos(30º) is positive or
negative. In either case, the other 2 angles are 150º, which
is in Quadrant 2, and 210º, which is in Quadrant 3.

Thus, all the angles x, are 30º, 150º, 210º and 330º.

In terms of π, knowing that π radians =180º, the
angles in radians are π/6, 5π/6, 7π/6 and 11π/6.

These are the only angles between 0º and 360º,
or, between 0 radians and 2π radians.

Answer 2

Make (cos x)^2 the subject by dividing by 4, then take the square root of both sides, remembering to put +/-.

You should be able to do this yourself and get
cos x = + or - (sqrt3)/2

In the first quadrant, the angle whose cos is (sqrt3)/2 is pi/6.

Since cos x can be either + or -, x can be in any of the four quadrants, and so as well as pi/6 you give the angle which corresponds to this in each of the other three quadrants. I expect you know how to do this and I don't want to spoil your fun by doing it for you.

P.S. Notice neither of the next 2 answerers is quite right: Helmut has done it in degrees instead of radians, and the next one ignores the negative value of cos x.

Answer 3

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Answer 4

cos^2(x) = 3/4
cos(x) = ±√0.75
x = π/6, 5π/6 , 7π/6, 11π/6 or
x = 30°, 150°, 210°, 330°

Answer 5

cos^2 x = 3 / 4
cos x = (sqrt 3) / 2
x = pi/6 , 11pi/6