Question

How do you find the general solutions for `3sec^2x + 4cos^2x = 7`?

Answer 1

Solve 3sec^2 x + 4cos^2 x = 7

Explanation:

Replace `sec^2 x` by `1/cos^2 x`
`3/cos^2 x + 4cos^2 x = 7`
`4cos^4 x - 7cos^2 x + 3 = 0` (Condition cos^2 x not zero)
Cal `cos^2 x = t`. We get a quadratic equation in t.
`4t^2 - 7t + 3 = 0.`
Since (a + b + c = 0), use the Shortcut. The 2 real roots are t = 1 and `t = 3/4.`
`t = cos^2 x = 1` --> `cos x = +- 1`--> `x = 0, x = pi, and x = 2pi`

`t = cos^2 x = 3/4` --> `cos x = +- sqrt3/2`

a. `cos x = sqrt3/2` --> `x = +- pi/6`
b. `cos x = -sqrt3/2`--> `x = +- (5pi)/6`