Calculus Question! Solve for x

Question

tan(x) + x(sec(x))² = 0

Answer 1

sec² x ((tan x / sec² x) + x) = 0

tan x / sec² x = (sin x / cos x) / (1 + tan² x)

= (sin x / cos x) / (1 + (sin² x / cos² x))

= (sin x / cos x) / (1 / cos² x)

= (sin x / cos x)

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

= (1/2) sin 2x

sec² x ((1/2) sin 2x + x) = 0

Can you solve using Zero Product Rule?

Answer 2

1+tan^2(x)= sec^2(x)

tanx + xsec^2(x) = 0

tanx +x(1+tan^2(x)) = 0

tanx +x + xtan^2(x) = 0

let y = tanx

xy^2 +y +x =0

y = -1/2x +/- (1/2x)sqr(1-4x^2)

x = tan^-1(y)

or

x(y^2+1)=-y

x= -y/(y^2+1) =-tanx/(1+tan^2(x)) = -tanx/sec^2(x)

= -sinx/cosx)(cos^2x) =-sinxcosx

= (1/2)sin2x

sin2x =2x

2x=0

x=0/2

x=0 whether measured in radians or degrees

although the tangent of any angle is a pure number, not in degrees or radians

so, technically there is no solution

but

ignore the unit problem and

check the answer

plug x=0 into

tanx +x(sec^2(x)=0

tan0 +0(sec^2(0))=0

0+0(1)=0

0=0