Question

Find all solutions of the equation in the interval [0, 2pi) tan^2(x)=sec(x)-1

Answer 1

Rewrite the equation recalling the definitions of secant and tangent:

$\dfrac{\sin^2(x)}{\cos^2(x)}=\dfrac{1}{\cos(x)}-1$

Rearrange the right hand side:

$\dfrac{\sin^2(x)}{\cos^2(x)}=\dfrac{1-\cos(x)}{\cos(x)}$

Multiply both sides by cos^2(x):

$\sin^2(x)=\cos(x)-\cos^2(x)$

Use the fundamental equation of trigonometry to express $\sin^2(x)$ in terms of $\cos^2(x)$:

$\cos^2(x)+\sin^2(x)=1 \iff \sin^2(x)=1-\cos^2(x)$

So, the expression becomes

$1-\cos^2(x)=\cos(x)-\cos^2(x)$

Simplifiy $-\cos^2(x)$ from both sides:

$1=\cos(x)$

Which only happens if x=0.