2tanx-sec^2 x=0 0<x<2π?

Question

Can anyone explain how to do this. I would be very very happy. Thank you.

Answer 1

2tanx - sec^2 x = 0

=> re-write sec^2 x as 1 + tan^2 x (that's an identity)

2tanx - 1 - tan^2 x = 0

=> let T = tanx

2T - 1 - T^2 = 0

=>

T^2 - 2T + 1 = 0

=> factor

(T - 1)^2 = 0

=>

T = 1

=>

tanx = 1

=> x = pi/4, 5pi/4

Answer 2

Let's deal with the second part:

0 For example, Sin(pi/2) = Sin(2*pi + pi/2) = Sin(4*pi + pi/2) and so on.

Now, for the main part:
2tanx-sec^2 x=0

One trick to solve differences worth 0, is to understand that the two values must be the same (otherwise, one minus the other would not be 0)

Then, you can use "identities" meaning trig functions that are worth the same as the ones you are given. You can do it directly by finding a relationship between tan and sec (and there is one -- see the answers above).

Or you can bring everything back to some basic function(s) like sin and cos.
tan = sin/cos
sec = 1/cos
sin(2x) = 2 sinx cosx

Here we go:
2tanx-sec^2 x=0
2 tanx = sec^2 x
2 sin/cos = 1 / cos^2 (I've dropped the x to simplify typing)
multiply the left side by cos/cos (this is a multiplication by 1 so it is OK)
2 sincos / cos^2 = 1/ cos^2
multiply both sides by cos^2
2 sincos = 1
sin(2x) = 1
2x = arcsin(1) = pi/2 (same as 90 degrees)
also (because sin is cyclical)
2x = arcsin(1) = 2 pi + pi/2
therefore
x = half of pi/2 = pi/4
and
x = half of (2*pi + pi/2) = pi + pi/4 (= 225 degrees)

Always check afterwards, just to make sure you (or I) did not make a mistake.

2 tan(pi/4) - sec^2(pi/4) = 2 - 2 = 0
2 tan(pi + pi/4) - sec^2(pi + pi/4) = 2 - 2 = 0

(The trap in the second one is that the secant is negative, but becomes positive when squared; proceed with caution).

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The "sin and cos" recipe takes longer but it should work in almost all cases. You'd have less identities to remember (and believe me, as you grow old and discover strange trig functions like haversines, you'll be happy to reduce everything to sin and cos).

Answer 3

Use the identity between tan and sec to express the LHS using only tan or sec. Then, it becomes a quadratic equation. Likely it can be solved with real roots for tanx or secx. From there, you can easily find the values for x.

Answer 4

2tanx-sec^2x=0
multiply both sides by cos^x;
2sinx cosx-1=0
sin(2x)=1 =sin(90).
2x=90
x=45 degrees.