What is the general solution for sec(t)= 1/(-(sqrt(2))/2) ?

The general solution for sec(t)= 1/(-(sqrt(2))/2) is t=(3pi)/4+2pin,t=(5pi)/4+2pin

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sec(t)= 1/(-(sqrt(2))/2)

sec (t) = \frac{1}{- \frac{2}{2}}

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

Solution steps

sec (t) = \frac{1}{- \frac{2}{2}}

Simplify

\frac{1}{- \frac{2}{2}} : - 2

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

Apply the fraction rule:

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

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

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

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

= - \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

General solutions for

sec (t) = - 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}

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

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

sin (x) = 0.556692

tan^{2} (x) + 4 tan (x) + 3 = 0

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

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

cot (x) = 1.2

What is the general solution for sec(t)= 1/(-(sqrt(2))/2) ?
The general solution for sec(t)= 1/(-(sqrt(2))/2) is t=(3pi)/4+2pin,t=(5pi)/4+2pin