521,417 views
23 votes
23 votes
Solve the equation for exact solutions over the interval [0, 2 π).sec^2x-2=tan^2x

User Emir Dupovac
by
2.9k points

1 Answer

12 votes
12 votes

sec^2 (x) -2 = tan^2 (x)

Rewritng sec and tan in terms of sin and cos

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

Multiply each term by cos^2(x)

1 - 2 cos^2(x) = sin^2(x)

Add 2 cos^2(x) to each side

1 = sin^2(x) + 2cos^2(x)

Break 2cos^2(x) into cos^2 (x) + cos^(x)

1 = sin^2(x) + cos^2(x) + cos^2(x)

We know that sin^2(x) + cos^2(x) = 1

1 = 1+cos^2(x)

Subtract 1 from each side

0 = cos^2(x)

Take the square root of each side

0 = cos(x)

x = pi/2, 3pi/2

But since we took the square root, we need to verify the solutions in the original equation

tan(x), sec^2 (x) cannot have cos(x) be zero so these solutions will not work

There are no solutions

No solutions

User Arkay Mann
by
3.1k points