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

The general solution for 2cos(x)=sec(x) is x= pi/4+2pin,x=(7pi)/4+2pin,x=(3pi)/4+2pin,x=(5pi)/4+2pin

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2cos(x)=sec(x)

Solution

2 cos (x) = sec (x)

Solution

x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

Degrees

x = 4 5^{\circ} + 36 0^{\circ} n, x = 31 5^{\circ} + 36 0^{\circ} n, x = 13 5^{\circ} + 36 0^{\circ} n, x = 22 5^{\circ} + 36 0^{\circ} n

Solution steps

2 cos (x) = sec (x)

Subtract

sec (x)

from both sides

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

Rewrite using trig identities

- sec (x) + 2 cos (x)

Use the basic trigonometric identity:

cos (x) = \frac{1}{sec ( x )}

= - sec (x) + 2 \cdot \frac{1}{sec ( x )}

2 \cdot \frac{1}{sec ( x )} = \frac{2}{sec ( x )}

2 \cdot \frac{1}{sec ( x )}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2}{sec ( x )}

Multiply the numbers:

1 \cdot 2 = 2

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

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

\frac{2}{sec ( x )} - sec (x) = 0

Solve by substitution

\frac{2}{sec ( x )} - sec (x) = 0

Let:

sec (x) = u

\frac{2}{u} - u = 0

\frac{2}{u} - u = 0 : u = 2, u = - 2

\frac{2}{u} - u = 0

Multiply both sides by

u

\frac{2}{u} - u = 0

Multiply both sides by

u

\frac{2}{u} u - uu = 0 \cdot u

Simplify

\frac{2}{u} u - uu = 0 \cdot u

Simplify

\frac{2}{u} u : 2

\frac{2}{u} u

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{2 u}{u}

Cancel the common factor:

u

= 2

Simplify

- uu : - u^{2}

- uu

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

uu = u^{1 + 1}

= - u^{1 + 1}

Add the numbers:

1 + 1 = 2

= - u^{2}

Simplify

0 \cdot u : 0

0 \cdot u

Apply rule

0 \cdot a = 0

= 0

2 - u^{2} = 0

2 - u^{2} = 0

2 - u^{2} = 0

Solve

2 - u^{2} = 0 : u = 2, u = - 2

2 - u^{2} = 0

Move

2

to the right side

2 - u^{2} = 0

Subtract

2

from both sides

2 - u^{2} - 2 = 0 - 2

Simplify

- u^{2} = - 2

- u^{2} = - 2

Divide both sides by

- 1

- u^{2} = - 2

Divide both sides by

- 1

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

Simplify

u^{2} = 2

u^{2} = 2

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = 2, u = - 2

u = 2, u = - 2

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

\frac{2}{u} - u

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = 2, u = - 2

Substitute back

u = sec (x)

sec (x) = 2, sec (x) = - 2

sec (x) = 2, sec (x) = - 2

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

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

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

sec (x) = - 2 : x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

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

Combine all the solutions

x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 2cos(x)=sec(x) ?
The general solution for 2cos(x)=sec(x) is x= pi/4+2pin,x=(7pi)/4+2pin,x=(3pi)/4+2pin,x=(5pi)/4+2pin