Question

Please please help!!

Answer 1

`\large\boxed{x=0\ and\ x=\pi}`

Explanation:

`\tan^2x\sec^2x+2\sec^2x-\tan^2x=2\\\\\text{Use}\ \tan x=(\sin x)/(\cos x),\ \sec x=(1)/(\cos x):\\\\\left((\sin x)/(\cos x)\right)^2\left((1)/(\cos x)\right)^2+2\left((1)/(\cos x)\right)^2-\left((\sin x)/(\cos x)\right)^2=2\\\\\left((\sin^2x)/(\cos^2x)\right)\left((1)/(\cos^2x)\right)+(2)/(\cos^2x)-(\sin^2x)/(\cos^2x)=2`

`(\sin^2x)/((\cos^2x)^2)+(2-\sin^2x)/(\cos^2x)=2\\\\\text{Use}\ \sin^2x+\cos^2x=1\to\sin^2x=1-\cos^2x\\\\(1-\cos^2x)/((\cos^2x)^2)+(2-(1-\cos^2x))/(\cos^2x)=2\\\\(1-\cos^2x)/((\cos^2x)^2)+(2-1+\cos^2x)/(\cos^2x)=2\\\\(1-\cos^2x)/((\cos^2x)^2)+(1+\cos^2x)/(\cos^2x)=2`

`(1-\cos^2x)/((\cos^2x)^2)+((1+\cos^2x)(\cos^2x))/((\cos^2x)^2)=2\qquad\text{Use the distributive property}\\\\(1-\cos^2x+\cos^2x+\cos^4x)/(\cos^4x)=2\\\\(1+\cos^4x)/(\cos^4x)=2\qquad\text{multiply both sides by}\ \cos^4x\\eq0\\\\1+\cos^4x=2\cos^4x\qquad\text{subtract}\ \cos^4x\ \text{from both sides}\\\\1=\cos^4x\iff \cos x=\pm\sqrt1\to\cos x=\pm1\\\\ x=k\pi\ for\ k\in\mathbb{Z}\\\\\text{On the interval}\ 0\leq x<2\pi,\ \text{the solutions are}\ x=0\ \text{and}\ x=\pi.`