What is the general solution for 1-1tan^2(2x)=sec^2(2x) ?

The general solution for 1-1tan^2(2x)=sec^2(2x) is x=(pin)/2

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

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

x = \frac{πn}{2}

Degrees

x = 0^{\circ} + 9 0^{\circ} n

Solution steps

1 - 1 \cdot tan^{2} (2 x) = sec^{2} (2 x)

Subtract

sec^{2} (2 x)

from both sides

1 - tan^{2} (2 x) - sec^{2} (2 x) = 0

Rewrite using trig identities

1 - sec^{2} (2 x) - tan^{2} (2 x)

Use the Pythagorean identity:

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

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

= - tan^{2} (2 x) - tan^{2} (2 x)

Simplify

= - 2 tan^{2} (2 x)

- 2 tan^{2} (2 x) = 0

Divide both sides by

- 2

- 2 tan^{2} (2 x) = 0

Divide both sides by

- 2

- 2 tan^{2} (2 x) = 0

Divide both sides by

- 2

\frac{- 2 tan ^{2} ( 2 x )}{- 2} = \frac{0}{- 2}

Simplify

tan^{2} (2 x) = 0

tan^{2} (2 x) = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

tan (2 x) = 0

General solutions for

tan (2 x) = 0

tan (x)

periodicity table with

πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

2 x = 0 + πn

2 x = 0 + πn

Solve

2 x = 0 + πn : x = \frac{πn}{2}

2 x = 0 + πn

0 + πn = πn

2 x = πn

Divide both sides by

2

2 x = πn

Divide both sides by

2

\frac{2 x}{2} = \frac{πn}{2}

Simplify

x = \frac{πn}{2}

x = \frac{πn}{2}

x = \frac{πn}{2}

10 tan (θ) - 4.9 (\frac{10}{13 cos ( θ )})^{2} = 0

1 - cos^{2} (θ) - 6959 cos (θ) = 0

(\frac{0.87 \cdot 1.04 \cdot sin ( 2 0 ^{\circ} - x )}{0.4}) = 0.05

sin (y) = 0, x cos (y) = 0

sin (6 0^{\circ} - x) - sin (6 0^{\circ} + x) = \frac{3}{2}

What is the general solution for 1-1tan^2(2x)=sec^2(2x) ?
The general solution for 1-1tan^2(2x)=sec^2(2x) is x=(pin)/2