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

1 Answer
Nghi N.
Aug 29, 2015

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

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