What is the general solution for 3cos(θ)+2=5sec(θ),0<= θ<2pi ?

Q: What is the general solution for 3cos(θ)+2=5sec(θ),0<= θ<2pi ?

The general solution for 3cos(θ)+2=5sec(θ),0<= θ<2pi is θ=0

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3cos(θ)+2=5sec(θ),0<= θ<2pi

Solution

3 cos (θ) + 2 = 5 sec (θ), 0 \leq θ < 2 π

Solution

θ = 0

Degrees

θ = 0^{\circ}

Solution steps

3 cos (θ) + 2 = 5 sec (θ), 0 \leq θ < 2 π

Subtract

5 sec (θ)

from both sides

3 cos (θ) + 2 - 5 sec (θ) = 0

Rewrite using trig identities

2 + 3 cos (θ) - 5 sec (θ)

Use the basic trigonometric identity:

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

= 2 + 3 \cdot \frac{1}{sec ( θ )} - 5 sec (θ)

3 \cdot \frac{1}{sec ( θ )} = \frac{3}{sec ( θ )}

3 \cdot \frac{1}{sec ( θ )}

Multiply fractions:

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

= \frac{1 \cdot 3}{sec ( θ )}

Multiply the numbers:

1 \cdot 3 = 3

= \frac{3}{sec ( θ )}

= 2 + \frac{3}{sec ( θ )} - 5 sec (θ)

2 + \frac{3}{sec ( θ )} - 5 sec (θ) = 0

Solve by substitution

2 + \frac{3}{sec ( θ )} - 5 sec (θ) = 0

Let:

sec (θ) = u

2 + \frac{3}{u} - 5 u = 0

2 + \frac{3}{u} - 5 u = 0 : u = - \frac{3}{5}, u = 1

2 + \frac{3}{u} - 5 u = 0

Multiply both sides by

u

2 + \frac{3}{u} - 5 u = 0

Multiply both sides by

u

2 u + \frac{3}{u} u - 5 uu = 0 \cdot u

Simplify

2 u + \frac{3}{u} u - 5 uu = 0 \cdot u

Simplify

\frac{3}{u} u : 3

\frac{3}{u} u

Multiply fractions:

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

= \frac{3 u}{u}

Cancel the common factor:

u

= 3

Simplify

- 5 uu : - 5 u^{2}

- 5 uu

Apply exponent rule:

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

uu = u^{1 + 1}

= - 5 u^{1 + 1}

Add the numbers:

1 + 1 = 2

= - 5 u^{2}

Simplify

0 \cdot u : 0

0 \cdot u

Apply rule

0 \cdot a = 0

= 0

2 u + 3 - 5 u^{2} = 0

2 u + 3 - 5 u^{2} = 0

2 u + 3 - 5 u^{2} = 0

Solve

2 u + 3 - 5 u^{2} = 0 : u = - \frac{3}{5}, u = 1

2 u + 3 - 5 u^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

- 5 u^{2} + 2 u + 3 = 0

Solve with the quadratic formula

- 5 u^{2} + 2 u + 3 = 0

Quadratic Equation Formula:

For

a = - 5, b = 2, c = 3

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 5 ) \cdot 3}{2 ( - 5 )}

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 5 ) \cdot 3}{2 ( - 5 )}

2^{2} - 4 (- 5) \cdot 3 = 8

2^{2} - 4 (- 5) \cdot 3

Apply rule

- (- a) = a

= 2^{2} + 4 \cdot 5 \cdot 3

Multiply the numbers:

4 \cdot 5 \cdot 3 = 60

= 2^{2} + 60

2^{2} = 4

= 4 + 60

Add the numbers:

4 + 60 = 64

= 64

Factor the number:

64 = 8^{2}

= 8^{2}

Apply radical rule:

8^{2} = 8

= 8

u_{1, 2} = \frac{- 2 \pm 8}{2 ( - 5 )}

Separate the solutions

u_{1} = \frac{- 2 + 8}{2 ( - 5 )}, u_{2} = \frac{- 2 - 8}{2 ( - 5 )}

u = \frac{- 2 + 8}{2 ( - 5 )} : - \frac{3}{5}

\frac{- 2 + 8}{2 ( - 5 )}

Remove parentheses:

(- a) = - a

= \frac{- 2 + 8}{- 2 \cdot 5}

Add/Subtract the numbers:

- 2 + 8 = 6

= \frac{6}{- 2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{6}{- 10}

Apply the fraction rule:

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

= - \frac{6}{10}

Cancel the common factor:

2

= - \frac{3}{5}

u = \frac{- 2 - 8}{2 ( - 5 )} : 1

\frac{- 2 - 8}{2 ( - 5 )}

Remove parentheses:

(- a) = - a

= \frac{- 2 - 8}{- 2 \cdot 5}

Subtract the numbers:

- 2 - 8 = - 10

= \frac{- 10}{- 2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{- 10}{- 10}

Apply the fraction rule:

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

= \frac{10}{10}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

u = - \frac{3}{5}, u = 1

u = - \frac{3}{5}, u = 1

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

2 + \frac{3}{u} - 5 u

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = - \frac{3}{5}, u = 1

Substitute back

u = sec (θ)

sec (θ) = - \frac{3}{5}, sec (θ) = 1

sec (θ) = - \frac{3}{5}, sec (θ) = 1

sec (θ) = - \frac{3}{5}, 0 \leq θ < 2 π :

No Solution

sec (θ) = - \frac{3}{5}, 0 \leq θ < 2 π

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

No Solution

sec (θ) = 1, 0 \leq θ < 2 π : θ = 0

sec (θ) = 1, 0 \leq θ < 2 π

General solutions for

sec (θ) = 1

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 = 0 + 2 πn

θ = 0 + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn

Solutions for the range

0 \leq θ < 2 π

θ = 0

Combine all the solutions

θ = 0

Popular Examples

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3 sec^{2} (x) = sec (x)

tan^2(x)-1=sec^2(x)

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

cos(3x)= pi/6

cos (3 x) = \frac{π}{6}

3-2sin^2(x)+5sin(x)=0

3 - 2 sin^{2} (x) + 5 sin (x) = 0

Frequently Asked Questions (FAQ)

What is the general solution for 3cos(θ)+2=5sec(θ),0<= θ<2pi ?
The general solution for 3cos(θ)+2=5sec(θ),0<= θ<2pi is θ=0