What is the general solution for 2sec^2(x)tan(x)-2sec^2(x)=0 ?

The general solution for 2sec^2(x)tan(x)-2sec^2(x)=0 is x= pi/4+pin

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

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

x = \frac{π}{4} + πn

Degrees

x = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

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

Factor

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

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

Factor out common term

2 sec^{2} (x)

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

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

Solving each part separately

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

sec^{2} (x) = 0 :

No Solution

sec^{2} (x) = 0

Apply rule

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

sec (x) = 0

sec (x) \leq - 1 or sec (x) \geq 1

No Solution

tan (x) - 1 = 0 : x = \frac{π}{4} + πn

tan (x) - 1 = 0

Move

1

to the right side

tan (x) - 1 = 0

Add

1

to both sides

tan (x) - 1 + 1 = 0 + 1

Simplify

tan (x) = 1

tan (x) = 1

General solutions for

tan (x) = 1

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}

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

Combine all the solutions

x = \frac{π}{4} + πn

sin (x) = \frac{10}{18}

tan (2 x) = 2 cos (x)

3 sec^{2} (θ) - 2 tan^{2} (θ) - 4 = 0

cos^{2} (x) + 5 sin^{2} (x) = 2

(sin (x))^{(s i n (x))} = \frac{1}{2}

What is the general solution for 2sec^2(x)tan(x)-2sec^2(x)=0 ?
The general solution for 2sec^2(x)tan(x)-2sec^2(x)=0 is x= pi/4+pin