Precalc math problem 2 and 3

Question

Solve the equation for exact solutions over the interval [0,2π).

1) sec² x + 1= 0

2) sec²x + 2sin x + 1= 0

Answer 1

sec^2(x)=-1

1/cos^2(x)=-1

cos^2(x)=-1

cosx =sqr(-1)= i,

x=cos^-1(i), which isn't possible

no solution for x, DNE, Does Not Exist, undefined

sec^2(x)+ 2sinx = -1

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

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

-1/(1-sin^2(x))= 1+sinx

1 +sinx -sin^2(x)-sin^3(x)= -1

let sinx =y, x=sin^-1(y)

y^3+y^2-y-2=0

1<y<2, but that's impossible, as all sines<1

x DNE, null set, no solution, undefined

or use a graphing calculator and

notice the curves never intersect the x axis, so no real solutions

Answer 2

Neither equation has a solution.

In equation 1, you cannot have a square equal to a negative value. sec²x cannot equal -1.

In equation 2, just type this into a graphing calculator and you will see that it never is equal to zero.