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

The general solution for sec(x)(2cos(x)-sqrt(2))=0 is x= pi/4+2pin,x=(7pi)/4+2pin

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sec(x)(2cos(x)-sqrt(2))=0

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

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

Degrees

x = 4 5^{\circ} + 36 0^{\circ} n, x = 31 5^{\circ} + 36 0^{\circ} n

Solution steps

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

Solving each part separately

sec (x) = 0 or 2 cos (x) - 2 = 0

sec (x) = 0 :

No Solution

sec (x) = 0

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

No Solution

2 cos (x) - 2 = 0 : x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

2 cos (x) - 2 = 0

Move

2

to the right side

2 cos (x) - 2 = 0

Add

2

to both sides

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

Simplify

2 cos (x) = 2

2 cos (x) = 2

Divide both sides by

2

2 cos (x) = 2

Divide both sides by

2

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

Simplify

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

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

General solutions for

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

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} 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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

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

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

Combine all the solutions

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

2 cos (θ) - 2 = 0

sin (x) - 2 = 0

x, sin^{4} (x) = \frac{cos ^{2} ( x )}{2}

tan^{2} (x) + 5 tan (x) - 1 = 0

csc (θ) + 1 = 0

What is the general solution for sec(x)(2cos(x)-sqrt(2))=0 ?
The general solution for sec(x)(2cos(x)-sqrt(2))=0 is x= pi/4+2pin,x=(7pi)/4+2pin