Question

sec(2x) = 3cos(2x) + cos(90 - 2x), - 90^0 < x < 90^0. Find x

Answer 1

We can transform this equation using the next trigonometric identities:

$$\sec x = \frac {1}{\cos x} \text{ and } \cos (90{}^{\circ} - x) = \sin x$$

We have:

$$\frac {1}{\cos (2 x)} = 3 \cos (2 x) + \sin (2 x)$$

Since $\sin x = \sqrt{1 - \cos^2 x}$ then:

$$\frac {1}{\cos (2 x)} = 3 \cos (2 x) + \sqrt {1 - \cos^ {2} (2 x)}$$

After the substitution $\cos (2x) = t$ we get the equation:

$$\frac {1}{t} = 3 t + \sqrt {1 - t ^ {2}}$$

The roots of this equation are: $-\frac{\sqrt{2}}{2}$ and $\frac{\sqrt{5}}{5}$

We have two equations for $\cos (2x)$:

$$\cos (2 x) = - \frac {\sqrt {2}}{2}$$ $$\cos (2 x) = \frac {\sqrt {5}}{5}$$

First equation gives $x = \pm \frac{1}{2}\arccos \left(-\frac{\sqrt{2}}{2}\right) = \pm 67.5{}^{\circ}$

From second equation $x = \pm \frac{1}{2}\arccos \left(-\frac{\sqrt{5}}{5}\right) \approx \pm 31.7{}^{\circ}$

Answer: $x = \{-67.5{}^{\circ}, -31.7{}^{\circ}, 31.7{}^{\circ}, 67.5{}^{\circ}\}$