# Solve for x the equation tan2x+1=1/cos2x if x is in interval (0,pi).

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### 1 Answer

First, we'll re-write the tangent function as the ratio sin 2x/cos 2x.

We'll re-write the equation:

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

We'll multiply by cos 2x all over:

sin 2x + cos 2x = 1

We'll recall the double angle identities for sin 2x and cos 2x:

sin 2x = 2 sinx*cos x

cos 2x = ` cos^2 x - sin^2 x`

We'll recall the Pythagorean identity:

`sin^2x + cos^2x = 1`

We'll re-write the equation:

`2sin x*cos x + cos^2 x - sin^2 x = sin^2 x + cos ^2 x`

We'll reduce like terms:

`2sin x*cos x - 2sin^2 x = 0`

We'll divide by 2 and we'll factor sin x:

sin x(cos x - sin x) = 0

We'll cancel each factor:

sin x = 0

x = 0, but the value doesn't belong to the interval (0,`pi` )

cos x - sin x = 0

We'll divide by sin x both sides:

cot x - 1 = 0

cot x = 1

The cotangent function has positive values within the interval (0;`pi`), only in the 1st quadrant.

x = `pi/4`

**Therefore,** **the only solution of the equation, over the interval `(0,pi)` is {`pi/4`}.**