Solve 3sec^2(x)-4cos^2(x)=7?

Question

Answers within the interval [0,2pi)

I got to [3/(cos^2(x)] - 4cos^2(x) = 7, now I have no idea how to factorise it for solving.

Answer 1

3 /cos²x - 4cos²x = 7
3 - 4.cos(^4)x = 7 cos²x
4 cos^(4) x + 7 cos² x - 3 = 0
Let y = cos²x
4y² + 7y - 3 = 0
y = [- 7 ± √97] / 8
y = 0.356 , y = - 2.1
cos² x = 0.356 is only possible solution
cos x = ± √ 0.356
cos x = ± 0.597
x = 0.93 , π - 0.93 , π + 0.93 , 2π - 0.93 radians
x = 0.93, 2.21, 4.07, 5.35 radians

Answer 2

Multiply by cos²x to get the following:

3 - 4cos^4 x = 7cos²x
OR
4cos^4 x + 7cos²x - 3 = 0

if it were +3, then this would factor. Since it doesn't, you can be creative with the quadratic formula, but you would be solving for cos²x since it's really a quartic that "looks" like a quadratic.

Answer 3

If you follow me, you'll get it.
Let y=cos (2x). Thus, your equation will be 3/y - 4y=7.

Multiply throughout by y. This means that every term in the equation you multiply it by y. The equation will be:
3-4y^2=7y

Rearrange the equation to something like this:
4y^2 + 7y + 3=0

Use your factorisation, to factorise the equation and you'll get this: (4y + 3)(y + 1)=0

(4y + 3) = 0 or (y + 1) =0
y = - 3/4 y = -1
cos 2x = -3/4 cos 2x = -1

After you get this, you find the reference angle and find at which quadrant the angle lies in.
Altogether, there's 3 angles/answers.

Happy Trying!

Answer 4

Now write it as
[ 3 - 4 cos^4 x - 7 cos^2 x] / cos^2 x = 0

This is zero when the numerator = 0

So let y = cos^2 x
and solve 3 - 7y - 4y^2 = 0