What is the general solution for sqrt(2)sec(x)+2=0 ?

The general solution for sqrt(2)sec(x)+2=0 is x=(3pi)/4+2pin,x=(5pi)/4+2pin

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sqrt(2)sec(x)+2=0

2 sec (x) + 2 = 0

x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

Degrees

x = 13 5^{\circ} + 36 0^{\circ} n, x = 22 5^{\circ} + 36 0^{\circ} n

Solution steps

2 sec (x) + 2 = 0

Move

2

to the right side

2 sec (x) + 2 = 0

Subtract

2

from both sides

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

Simplify

2 sec (x) = - 2

2 sec (x) = - 2

Divide both sides by

2

2 sec (x) = - 2

Divide both sides by

2

\frac{2 sec ( x )}{2} = \frac{- 2}{2}

Simplify

\frac{2 sec ( x )}{2} = \frac{- 2}{2}

Simplify

\frac{2 sec ( x )}{2} : sec (x)

\frac{2 sec ( x )}{2}

Cancel the common factor:

2

= sec (x)

Simplify

\frac{- 2}{2} : - 2

\frac{- 2}{2}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{2}{2}

Apply radical rule:

2 = 2^{\frac{1}{2}}

= \frac{2}{2 ^{\frac{1}{2}}}

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = x^{a - b}

\frac{2 ^{1}}{2 ^{\frac{1}{2}}} = 2^{1 - \frac{1}{2}}

= 2^{1 - \frac{1}{2}}

Subtract the numbers:

1 - \frac{1}{2} = \frac{1}{2}

= 2^{\frac{1}{2}}

Apply radical rule:

2^{\frac{1}{2}} = 2

= - 2

sec (x) = - 2

sec (x) = - 2

sec (x) = - 2

General solutions for

sec (x) = - 2

sec (x)

periodicity table with

2 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} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

6 sin (θ) - 3 = 0

cot^{2} (x) - csc (x) = 1

cos (2 x) = - 0.6

2 sin (x) - cos (x) = 0

2 cos (x) = sin (2 x)

What is the general solution for sqrt(2)sec(x)+2=0 ?
The general solution for sqrt(2)sec(x)+2=0 is x=(3pi)/4+2pin,x=(5pi)/4+2pin