What is the general solution for solvefor x,t^2=2sec(x) ?

The general solution for solvefor x,t^2=2sec(x) is x=arcsec((t^2)/2)+2pin,x=-arcsec((t^2)/2)+2pin

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solvefor x,t^2=2sec(x)

solve for

x, t^{2} = 2 sec (x)

x = arcsec (\frac{t ^{2}}{2}) + 2 πn, x = - arcsec (\frac{t ^{2}}{2}) + 2 πn

Solution steps

t^{2} = 2 sec (x)

Switch sides

2 sec (x) = t^{2}

Divide both sides by

2

2 sec (x) = t^{2}

Divide both sides by

2

\frac{2 sec ( x )}{2} = \frac{t ^{2}}{2}

Simplify

sec (x) = \frac{t ^{2}}{2}

sec (x) = \frac{t ^{2}}{2}

Apply trig inverse properties

sec (x) = \frac{t ^{2}}{2}

General solutions for

sec (x) = \frac{t ^{2}}{2}

sec (x) = a \Rightarrow x = arcsec (a) + 2 πn, x = - arcsec (a) + 2 πn

x = arcsec (\frac{t ^{2}}{2}) + 2 πn, x = - arcsec (\frac{t ^{2}}{2}) + 2 πn

x = arcsec (\frac{t ^{2}}{2}) + 2 πn, x = - arcsec (\frac{t ^{2}}{2}) + 2 πn

(8 cos (x) - 4) (sin (x)) = 0, 0 \leq x < 2 π

sin (x) = 3511

tan^{2} (x) - 1 = 4 cos (x)

\frac{10}{sin ( 3 0 ^{\circ} )} = \frac{3}{sin ( b )}

tan (\frac{θ}{2} - \frac{π}{6}) = - 1, 0 \leq θ < 2 π

What is the general solution for solvefor x,t^2=2sec(x) ?
The general solution for solvefor x,t^2=2sec(x) is x=arcsec((t^2)/2)+2pin,x=-arcsec((t^2)/2)+2pin