Find all solutions of the equation on the interval [0,2pi):

Tan^2x=1-secx

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solve tan^2(x)=1-sec(x)

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Simplify trigonometric functions:

If we know that:

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

tan^2(x)+sec^2(x)-1=0 becomes:

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

sec^2 x - 1= 1 - secx

sec^2 x + sec x - 2 = 0
(secx + 2)(secx -1) = 0
secx = -2 or secx = 1
cosx = -1/2 or cosx = 1

x = 120° or x = 240° or x = 0 or x = 360°

x = 0, 2π/3 , 4π/3, 2π

Thank you!

find all solutions to the equation 2x^3+12x^2+50x=0

To find all solutions of the equation tan^2(x) = 1 - sec(x) on the interval [0, 2π), we need to simplify the equation and then solve for x.

Let's start by simplifying the equation:
tan^2(x) = 1 - sec(x)
Using the identity: tan^2(x) = 1 - cos^2(x), we can rewrite the equation as:
1 - cos^2(x) = 1 - sec(x)

Next, let's express sec(x) in terms of cos(x):
sec(x) = 1/cos(x)

Substituting this into the equation, we get:
1 - cos^2(x) = 1 - 1/cos(x)

Now, to remove the fraction, let's multiply both sides of the equation by cos(x):
cos(x) - cos^3(x) = cos(x) - 1

Rearranging the terms, we have:
cos^3(x) = 1

Taking the cube root of both sides, we get:
cos(x) = 1

Now, solving for x:
x = 0

So, the equation has one solution on the interval [0, 2π), which is x = 0.