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

Question

asked Oct 7, 2020 94.9k views

1 Answer

Ashish Rathi · Answer 1 · 2020-10-14T15:53:54+0000

Rewrite the equation recalling the definitions of secant and tangent:

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

Rearrange the right hand side:

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

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.