How can I solve this for x? Please show your work...I'm extremely lost!?

Question

secxtanx((tanx)^2+(secx)^2))=0

Answer 1

Since you basically have 3 things multiplied together, it comes out 0 if any of the 3 equals zero. secx is never zero. tanx is zero whenever sinx = 0 since tan = sin/cos. This happens if x = 0, 180, 360, etc or 0, pi, 2pi,...

And the third can be written as sin^2 x/cos^2 x + 1/cos^2 x which is zero whenever the numerator adds to zero meaning sin^2 x would have to be -1 which is impossible. So only multiples of 180 or pi.

Answer 2

secxtanx((tanx)^2+(secx)^2))=0 when sec(x)=0, tan(x)=0, or (tanx)^2+(secx)^2)=0.

sec(x) is never equal to 0, and there is no value of x for which (tanx)^2+(secx)^2) is equal to zero.

Therefor the function is equal to 0 only when tan(x)=0. Therefore x=...,-2pi,-pi,0, pi, 2pi,...

Answer 3

Sorry you are so lost!
With a bit of work, this equation reduces
quite nicely.
First, sec x is never 0, since it is the
reciprocal of the cosine.
So we can divide it out.
Next, tan² x + sec² x = 1 + 2 tan² x.
If this were 0, we would get
tan² x = -1/2, which is impossible.
So all that's left is tan x = 0
and the solutions for 0 <= x <= 2π
are x = 0, π and 2π.

Answer 4

(tan x)^2 is denoted tan^2 x, and (sec x)^2 is denoted sec^2 x.

I will solve the equation for x in the interval [0, 2pi)

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

If any of the factors equal zero, the equation works, so

sec x = 0 or tan x = 0 or tan^2 x + sec^2 x = 0

I will start with sec x = 0.

Identity:
sec x = 1/cos x

(times each side by cos x)

1 = 0
this equation is never true.

So,
tan x = 0 or tan^2 x + sec^2 x = 0

tan x = 0

x = {0, pi}

Now I will solve for
tan^2 x + sec^2 x = 0

Pythagorean identity:
tan^2 x = sec^2 x - 1

sec^2 x - 1 + sec^2 x = 0

2sec^2 x = 1

sec^2 x = 1/2

sec x = +/- √(2)/2

Identity:
sec x = 1/cos x

1/cos x = +/- √(2)/2

cos x = +/- √(2)

cos x can never be greater than 1

answer:
x = {0, pi}

Answer 5

secxtanx((tanx)^2+(secx)^2))=0
1/cosx*sinx/cosx(sin^2x/cos^2x + 1 cos^2x) =0
[sinx/cos^2x][(sin^2x+1)/cos^2x)] = 0
sinx(sin^2x +1) = 0
sin x = 0 --> x = 0, pi, 2pi and multiples thereof
sin^x+1 =0
sin^2x = -1 which gives imaginary values for x