Question

Find all solutions of each equation on the interval 0 less than or equal to x less than 2pi

tan^2x sec^2x +2sec^2x-tan^2x=2​

Answer 1

`x=0,\pi,2\pi`

Explanation:

The given trigonometric equation is

`\tan^2(x) \sec^2(x)+2\sec^2(x)-\tan^2(x)=2`.

Let us equate everything to zero

`\tan^2(x) \sec^2(x)+2\sec^2(x)-\tan^2(x)-2=0`.

We factor now to get:

`\sec^2(x)(\tan^2(x)+2)-1(\tan^2(x)+2)=0`.

`(\sec^2(x)-1)(\tan^2(x)+2)=0`.

Apply zero product property:

`(\sec^2(x)-1)=0,(\tan^2(x)+2)=0`.

`\sec^2(x)=1,\tan^2(x)=-2`.

When

`\sec^2(x)=1`.

`\implies \sec(x)=\pm1`.

Reciprocate both sides to get;

`\implies \cos(x)=\pm1`.

`x=0,\pi,2\pi`

When
`\tan^2(x)=-2` we don't have real solutions.

Therefore
`x=0,\pi,2\pi` on
`0\le x\le 2\pi`.