What is the general solution for sec(θ)-sqrt(2)=0,0<= θ<= 2pi ?

The general solution for sec(θ)-sqrt(2)=0,0<= θ<= 2pi is θ= pi/4 ,θ=(7pi)/4

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sec(θ)-sqrt(2)=0,0<= θ<= 2pi

sec (θ) - 2 = 0, 0 \leq θ \leq 2 π

θ = \frac{π}{4}, θ = \frac{7 π}{4}

Degrees

θ = 4 5^{\circ}, θ = 31 5^{\circ}

Solution steps

sec (θ) - 2 = 0, 0 \leq θ \leq 2 π

Move

2

to the right side

sec (θ) - 2 = 0

Add

2

to both sides

sec (θ) - 2 + 2 = 0 + 2

Simplify

sec (θ) = 2

sec (θ) = 2

General solutions for

sec (θ) = 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}

θ = \frac{π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

θ = \frac{π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

Solutions for the range

0 \leq θ \leq 2 π

θ = \frac{π}{4}, θ = \frac{7 π}{4}

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

cos (θ) = \frac{7}{3 17}

sin (A) - 0.1 \cdot cos (A) = \frac{6.94}{9.8}

t, s = 2 arctan (t)

\frac{cos ( x )}{tan ( x )} = \frac{3}{2}

What is the general solution for sec(θ)-sqrt(2)=0,0<= θ<= 2pi ?
The general solution for sec(θ)-sqrt(2)=0,0<= θ<= 2pi is θ= pi/4 ,θ=(7pi)/4