Question

Find all solutions of each equation on the intercal 0 <_ x < 2π

tan ²x.sec²x -2sec²-tan²x=2

Answer 1

To find the solutions of the equation tan²x.sec²x - 2sec²x - tan²x = 2, we can simplify the equation and make a substitution. By factoring the equation and solving for the possible solutions, we find that the only solution is u = 2.

Step-by-step explanation:

To find the solutions of the equation tan²x.sec²x - 2sec²x - tan²x = 2

We can start by simplifying the equation:

tan²x.sec²x - 2sec²x - tan²x = 2

tan²x(sec²x - 1) - sec²x = 2

tan²x.sec²x - tan²x - sec²x - 2 = 0

Now, we can make a substitution: u = tan²x.sec²x

Using this substitution, the equation becomes: u - tan²x - sec²x - 2 = 0

Next, we can factor the equation: (u - 2)(tan²x + sec²x + 1) = 0

From this, we can see that there are two possible solutions:

u - 2 = 0 (which gives us u = 2)

or

tan²x + sec²x + 1 = 0

To solve tan²x + sec²x + 1 = 0, we can make another substitution: v = tanx + secx

Using this substitution, the equation becomes: v² + 1 = 0

Since v² + 1 = 0 has no real solutions, we can conclude that the only solution to the original equation is u = 2.