What is the general solution for 5sec^2(θ)sin(θ)-cos(θ)=0 ?

The general solution for 5sec^2(θ)sin(θ)-cos(θ)=0 is θ=0.19049…+2pin,θ=-2.95110…+2pin

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Popular Trigonometry >

5sec^2(θ)sin(θ)-cos(θ)=0

Solution

5 sec^{2} (θ) sin (θ) - cos (θ) = 0

Solution

θ = 0.19049 \dots + 2 πn, θ = - 2.95110 \dots + 2 πn

Degrees

θ = 10.91437 \dots^{\circ} + 36 0^{\circ} n, θ = - 169.08562 \dots^{\circ} + 36 0^{\circ} n

Solution steps

5 sec^{2} (θ) sin (θ) - cos (θ) = 0

Express with sin, cos

5 (\frac{1}{cos ( θ )})^{2} sin (θ) - cos (θ) = 0

Simplify

5 (\frac{1}{cos ( θ )})^{2} sin (θ) - cos (θ) : \frac{5 sin ( θ ) - cos ^{3} ( θ )}{cos ^{2} ( θ )}

5 (\frac{1}{cos ( θ )})^{2} sin (θ) - cos (θ)

5 (\frac{1}{cos ( θ )})^{2} sin (θ) = \frac{5 sin ( θ )}{cos ^{2} ( θ )}

5 (\frac{1}{cos ( θ )})^{2} sin (θ)

(\frac{1}{cos ( θ )})^{2} = \frac{1}{cos ^{2} ( θ )}

(\frac{1}{cos ( θ )})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{2}}{cos ^{2} ( θ )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{cos ^{2} ( θ )}

= 5 \cdot \frac{1}{cos ^{2} ( θ )} sin (θ)

Multiply fractions:

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

= \frac{1 \cdot 5 sin ( θ )}{cos ^{2} ( θ )}

Multiply the numbers:

1 \cdot 5 = 5

= \frac{5 sin ( θ )}{cos ^{2} ( θ )}

= \frac{5 sin ( θ )}{cos ^{2} ( θ )} - cos (θ)

Convert element to fraction:

cos (θ) = \frac{cos ( θ ) cos ^{2} ( θ )}{cos ^{2} ( θ )}

= \frac{5 sin ( θ )}{cos ^{2} ( θ )} - \frac{cos ( θ ) cos ^{2} ( θ )}{cos ^{2} ( θ )}

Since the denominators are equal, combine the fractions:

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

= \frac{5 sin ( θ ) - cos ( θ ) cos ^{2} ( θ )}{cos ^{2} ( θ )}

5 sin (θ) - cos (θ) cos^{2} (θ) = 5 sin (θ) - cos^{3} (θ)

5 sin (θ) - cos (θ) cos^{2} (θ)

cos (θ) cos^{2} (θ) = cos^{3} (θ)

cos (θ) cos^{2} (θ)

Apply exponent rule:

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

cos (θ) cos^{2} (θ) = cos^{1 + 2} (θ)

= cos^{1 + 2} (θ)

Add the numbers:

1 + 2 = 3

= cos^{3} (θ)

= 5 sin (θ) - cos^{3} (θ)

= \frac{5 sin ( θ ) - cos ^{3} ( θ )}{cos ^{2} ( θ )}

\frac{5 sin ( θ ) - cos ^{3} ( θ )}{cos ^{2} ( θ )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

5 sin (θ) - cos^{3} (θ) = 0

Add

cos^{3} (θ)

to both sides

5 sin (θ) = cos^{3} (θ)

Square both sides

(5 sin (θ))^{2} = (cos^{3} (θ))^{2}

Subtract

(cos^{3} (θ))^{2}

from both sides

25 sin^{2} (θ) - cos^{6} (θ) = 0

Rewrite using trig identities

- cos^{6} (θ) + 25 sin^{2} (θ)

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

sin^{2} (x) = 1 - cos^{2} (x)

= - cos^{6} (θ) + 25 (1 - cos^{2} (θ))

- cos^{6} (θ) + (1 - cos^{2} (θ)) \cdot 25 = 0

Solve by substitution

- cos^{6} (θ) + (1 - cos^{2} (θ)) \cdot 25 = 0

Let:

cos (θ) = u

- u^{6} + (1 - u^{2}) \cdot 25 = 0

- u^{6} + (1 - u^{2}) \cdot 25 = 0 : u = 0.96414 \dots, u = - 0.96414 \dots

- u^{6} + (1 - u^{2}) \cdot 25 = 0

Expand

- u^{6} + (1 - u^{2}) \cdot 25 : - u^{6} + 25 - 25 u^{2}

- u^{6} + (1 - u^{2}) \cdot 25

= - u^{6} + 25 (1 - u^{2})

Expand

25 (1 - u^{2}) : 25 - 25 u^{2}

25 (1 - u^{2})

Apply the distributive law:

a (b - c) = ab - a c

a = 25, b = 1, c = u^{2}

= 25 \cdot 1 - 25 u^{2}

Multiply the numbers:

25 \cdot 1 = 25

= 25 - 25 u^{2}

= - u^{6} + 25 - 25 u^{2}

- u^{6} + 25 - 25 u^{2} = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

- u^{6} - 25 u^{2} + 25 = 0

Rewrite the equation with

v = u^{2}

and

v^{3} = u^{6}

- v^{3} - 25 v + 25 = 0

Solve

- v^{3} - 25 v + 25 = 0 : v \approx 0.96414 \dots

- v^{3} - 25 v + 25 = 0

Find one solution for

- v^{3} - 25 v + 25 = 0

using Newton-Raphson:

v \approx 0.96414 \dots

- v^{3} - 25 v + 25 = 0

Newton-Raphson Approximation Definition

f (v) = - v^{3} - 25 v + 25

Find

f^{^{'}} (v) : - 3 v^{2} - 25

\frac{d}{d v} (- v^{3} - 25 v + 25)

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= - \frac{d}{d v} (v^{3}) - \frac{d}{d v} (25 v) + \frac{d}{d v} (25)

\frac{d}{d v} (v^{3}) = 3 v^{2}

\frac{d}{d v} (v^{3})

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 3 v^{3 - 1}

Simplify

= 3 v^{2}

\frac{d}{d v} (25 v) = 25

\frac{d}{d v} (25 v)

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 25 \frac{d v}{d v}

Apply the common derivative:

\frac{d v}{d v} = 1

= 25 \cdot 1

Simplify

= 25

\frac{d}{d v} (25) = 0

\frac{d}{d v} (25)

Derivative of a constant:

\frac{d}{d x} (a) = 0

= 0

= - 3 v^{2} - 25 + 0

Simplify

= - 3 v^{2} - 25

Let

v_{0} = 1

Compute

v_{n + 1}

until

Δ v_{n + 1} < 0.000001

v_{1} = 0.96428 \dots : Δ v_{1} = 0.03571 \dots

f (v_{0}) = - 1^{3} - 25 \cdot 1 + 25 = - 1

f^{^{'}} (v_{0}) = - 3 \cdot 1^{2} - 25 = - 28

v_{1} = 0.96428 \dots

Δ v_{1} = ∣ 0.96428 \dots - 1 ∣ = 0.03571 \dots

Δ v_{1} = 0.03571 \dots

v_{2} = 0.96414 \dots : Δ v_{2} = 0.00013 \dots

f (v_{1}) = - 0.96428 \dots^{3} - 25 \cdot 0.96428 \dots + 25 = - 0.00378 \dots

f^{^{'}} (v_{1}) = - 3 \cdot 0.96428 \dots^{2} - 25 = - 27.78954 \dots

v_{2} = 0.96414 \dots

Δ v_{2} = ∣ 0.96414 \dots - 0.96428 \dots ∣ = 0.00013 \dots

Δ v_{2} = 0.00013 \dots

v_{3} = 0.96414 \dots : Δ v_{3} = 1.927 E - 9

f (v_{2}) = - 0.96414 \dots^{3} - 25 \cdot 0.96414 \dots + 25 = - 5.35491 E - 8

f^{^{'}} (v_{2}) = - 3 \cdot 0.96414 \dots^{2} - 25 = - 27.78875 \dots

v_{3} = 0.96414 \dots

Δ v_{3} = ∣ 0.96414 \dots - 0.96414 \dots ∣ = 1.927 E - 9

Δ v_{3} = 1.927 E - 9

v \approx 0.96414 \dots

Apply long division:

\frac{- v ^{3} - 25 v + 25}{v - 0.96414 \dots} = - v^{2} - 0.96414 \dots v - 25.92958 \dots

- v^{2} - 0.96414 \dots v - 25.92958 \dots \approx 0

Find one solution for

- v^{2} - 0.96414 \dots v - 25.92958 \dots = 0

using Newton-Raphson:

No Solution for

v \in R

- v^{2} - 0.96414 \dots v - 25.92958 \dots = 0

Newton-Raphson Approximation Definition

f (v) = - v^{2} - 0.96414 \dots v - 25.92958 \dots

Find

f^{^{'}} (v) : - 2 v - 0.96414 \dots

\frac{d}{d v} (- v^{2} - 0.96414 \dots v - 25.92958 \dots)

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= - \frac{d}{d v} (v^{2}) - \frac{d}{d v} (0.96414 \dots v) - \frac{d}{d v} (25.92958 \dots)

\frac{d}{d v} (v^{2}) = 2 v

\frac{d}{d v} (v^{2})

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 2 v^{2 - 1}

Simplify

= 2 v

\frac{d}{d v} (0.96414 \dots v) = 0.96414 \dots

\frac{d}{d v} (0.96414 \dots v)

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 0.96414 \dots \frac{d v}{d v}

Apply the common derivative:

\frac{d v}{d v} = 1

= 0.96414 \dots \cdot 1

Simplify

= 0.96414 \dots

\frac{d}{d v} (25.92958 \dots) = 0

\frac{d}{d v} (25.92958 \dots)

Derivative of a constant:

\frac{d}{d x} (a) = 0

= 0

= - 2 v - 0.96414 \dots - 0

Simplify

= - 2 v - 0.96414 \dots

Let

v_{0} = - 5

Compute

v_{n + 1}

until

Δ v_{n + 1} < 0.000001

v_{1} = 0.10287 \dots : Δ v_{1} = 5.10287 \dots

f (v_{0}) = - (- 5)^{2} - 0.96414 \dots (- 5) - 25.92958 \dots = - 46.10883 \dots

f^{^{'}} (v_{0}) = - 2 (- 5) - 0.96414 \dots = 9.03585 \dots

v_{1} = 0.10287 \dots

Δ v_{1} = ∣ 0.10287 \dots - (- 5) ∣ = 5.10287 \dots

Δ v_{1} = 5.10287 \dots

v_{2} = - 22.15480 \dots : Δ v_{2} = 22.25767 \dots

f (v_{1}) = - 0.10287 \dots^{2} - 0.96414 \dots \cdot 0.10287 \dots - 25.92958 \dots = - 26.03935 \dots

f^{^{'}} (v_{1}) = - 2 \cdot 0.10287 \dots - 0.96414 \dots = - 1.16990 \dots

v_{2} = - 22.15480 \dots

Δ v_{2} = ∣ - 22.15480 \dots - 0.10287 \dots ∣ = 22.25767 \dots

Δ v_{2} = 22.25767 \dots

v_{3} = - 10.72559 \dots : Δ v_{3} = 11.42920 \dots

f (v_{2}) = - (- 22.15480 \dots)^{2} - 0.96414 \dots (- 22.15480 \dots) - 25.92958 \dots = - 495.40430 \dots

f^{^{'}} (v_{2}) = - 2 (- 22.15480 \dots) - 0.96414 \dots = 43.34545 \dots

v_{3} = - 10.72559 \dots

Δ v_{3} = ∣ - 10.72559 \dots - (- 22.15480 \dots) ∣ = 11.42920 \dots

Δ v_{3} = 11.42920 \dots

v_{4} = - 4.34951 \dots : Δ v_{4} = 6.37607 \dots

f (v_{3}) = - (- 10.72559 \dots)^{2} - 0.96414 \dots (- 10.72559 \dots) - 25.92958 \dots = - 130.62684 \dots

f^{^{'}} (v_{3}) = - 2 (- 10.72559 \dots) - 0.96414 \dots = 20.48703 \dots

v_{4} = - 4.34951 \dots

Δ v_{4} = ∣ - 4.34951 \dots - (- 10.72559 \dots) ∣ = 6.37607 \dots

Δ v_{4} = 6.37607 \dots

v_{5} = 0.90644 \dots : Δ v_{5} = 5.25596 \dots

f (v_{4}) = - (- 4.34951 \dots)^{2} - 0.96414 \dots (- 4.34951 \dots) - 25.92958 \dots = - 40.65431 \dots

f^{^{'}} (v_{4}) = - 2 (- 4.34951 \dots) - 0.96414 \dots = 7.73488 \dots

v_{5} = 0.90644 \dots

Δ v_{5} = ∣ 0.90644 \dots - (- 4.34951 \dots) ∣ = 5.25596 \dots

Δ v_{5} = 5.25596 \dots

v_{6} = - 9.04124 \dots : Δ v_{6} = 9.94769 \dots

f (v_{5}) = - 0.90644 \dots^{2} - 0.96414 \dots \cdot 0.90644 \dots - 25.92958 \dots = - 27.62518 \dots

f^{^{'}} (v_{5}) = - 2 \cdot 0.90644 \dots - 0.96414 \dots = - 2.77704 \dots

v_{6} = - 9.04124 \dots

Δ v_{6} = ∣ - 9.04124 \dots - 0.90644 \dots ∣ = 9.94769 \dots

Δ v_{6} = 9.94769 \dots

v_{7} = - 3.26050 \dots : Δ v_{7} = 5.78073 \dots

f (v_{6}) = - (- 9.04124 \dots)^{2} - 0.96414 \dots (- 9.04124 \dots) - 25.92958 \dots = - 98.95656 \dots

f^{^{'}} (v_{6}) = - 2 (- 9.04124 \dots) - 0.96414 \dots = 17.11833 \dots

v_{7} = - 3.26050 \dots

Δ v_{7} = ∣ - 3.26050 \dots - (- 9.04124 \dots) ∣ = 5.78073 \dots

Δ v_{7} = 5.78073 \dots

v_{8} = 2.75310 \dots : Δ v_{8} = 6.01361 \dots

f (v_{7}) = - (- 3.26050 \dots)^{2} - 0.96414 \dots (- 3.26050 \dots) - 25.92958 \dots = - 33.41688 \dots

f^{^{'}} (v_{7}) = - 2 (- 3.26050 \dots) - 0.96414 \dots = 5.55687 \dots

v_{8} = 2.75310 \dots

Δ v_{8} = ∣ 2.75310 \dots - (- 3.26050 \dots) ∣ = 6.01361 \dots

Δ v_{8} = 6.01361 \dots

v_{9} = - 2.83600 \dots : Δ v_{9} = 5.58911 \dots

f (v_{8}) = - 2.75310 \dots^{2} - 0.96414 \dots \cdot 2.75310 \dots - 25.92958 \dots = - 36.16359 \dots

f^{^{'}} (v_{8}) = - 2 \cdot 2.75310 \dots - 0.96414 \dots = - 6.47036 \dots

v_{9} = - 2.83600 \dots

Δ v_{9} = ∣ - 2.83600 \dots - 2.75310 \dots ∣ = 5.58911 \dots

Δ v_{9} = 5.58911 \dots

v_{10} = 3.79931 \dots : Δ v_{10} = 6.63532 \dots

f (v_{9}) = - (- 2.83600 \dots)^{2} - 0.96414 \dots (- 2.83600 \dots) - 25.92958 \dots = - 31.23817 \dots

f^{^{'}} (v_{9}) = - 2 (- 2.83600 \dots) - 0.96414 \dots = 4.70786 \dots

v_{10} = 3.79931 \dots

Δ v_{10} = ∣ 3.79931 \dots - (- 2.83600 \dots) ∣ = 6.63532 \dots

Δ v_{10} = 6.63532 \dots

Cannot find solution

The solution is

v \approx 0.96414 \dots

v \approx 0.96414 \dots

Substitute back

v = u^{2},

solve for

u

Solve

u^{2} = 0.96414 \dots : u = 0.96414 \dots, u = - 0.96414 \dots

u^{2} = 0.96414 \dots

For

x^{2} = f (a)

the solutions are

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

u = 0.96414 \dots, u = - 0.96414 \dots

The solutions are

u = 0.96414 \dots, u = - 0.96414 \dots

Substitute back

u = cos (θ)

cos (θ) = 0.96414 \dots, cos (θ) = - 0.96414 \dots

cos (θ) = 0.96414 \dots, cos (θ) = - 0.96414 \dots

cos (θ) = 0.96414 \dots : θ = arccos (0.96414 \dots) + 2 πn, θ = 2 π - arccos (0.96414 \dots) + 2 πn

cos (θ) = 0.96414 \dots

Apply trig inverse properties

cos (θ) = 0.96414 \dots

General solutions for

cos (θ) = 0.96414 \dots

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

θ = arccos (0.96414 \dots) + 2 πn, θ = 2 π - arccos (0.96414 \dots) + 2 πn

θ = arccos (0.96414 \dots) + 2 πn, θ = 2 π - arccos (0.96414 \dots) + 2 πn

cos (θ) = - 0.96414 \dots : θ = arccos (- 0.96414 \dots) + 2 πn, θ = - arccos (- 0.96414 \dots) + 2 πn

cos (θ) = - 0.96414 \dots

Apply trig inverse properties

cos (θ) = - 0.96414 \dots

General solutions for

cos (θ) = - 0.96414 \dots

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

θ = arccos (- 0.96414 \dots) + 2 πn, θ = - arccos (- 0.96414 \dots) + 2 πn

θ = arccos (- 0.96414 \dots) + 2 πn, θ = - arccos (- 0.96414 \dots) + 2 πn

Combine all the solutions

θ = arccos (0.96414 \dots) + 2 πn, θ = 2 π - arccos (0.96414 \dots) + 2 πn, θ = arccos (- 0.96414 \dots) + 2 πn, θ = - arccos (- 0.96414 \dots) + 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

5 sec^{2} (θ) sin (θ) - cos (θ) = 0

Remove the ones that don't agree with the equation.

Check the solution

arccos (0.96414 \dots) + 2 πn :

True

arccos (0.96414 \dots) + 2 πn

Plug in

n = 1

arccos (0.96414 \dots) + 2 π 1

For

5 sec^{2} (θ) sin (θ) - cos (θ) = 0

plug in

θ = arccos (0.96414 \dots) + 2 π 1

5 sec^{2} (arccos (0.96414 \dots) + 2 π 1) sin (arccos (0.96414 \dots) + 2 π 1) - cos (arccos (0.96414 \dots) + 2 π 1) = 0

Refine

0 = 0

\Rightarrow True

Check the solution

2 π - arccos (0.96414 \dots) + 2 πn :

False

2 π - arccos (0.96414 \dots) + 2 πn

Plug in

n = 1

2 π - arccos (0.96414 \dots) + 2 π 1

For

5 sec^{2} (θ) sin (θ) - cos (θ) = 0

plug in

θ = 2 π - arccos (0.96414 \dots) + 2 π 1

5 sec^{2} (2 π - arccos (0.96414 \dots) + 2 π 1) sin (2 π - arccos (0.96414 \dots) + 2 π 1) - cos (2 π - arccos (0.96414 \dots) + 2 π 1) = 0

Refine

- 1.96382 \dots = 0

\Rightarrow False

Check the solution

arccos (- 0.96414 \dots) + 2 πn :

False

arccos (- 0.96414 \dots) + 2 πn

Plug in

n = 1

arccos (- 0.96414 \dots) + 2 π 1

For

5 sec^{2} (θ) sin (θ) - cos (θ) = 0

plug in

θ = arccos (- 0.96414 \dots) + 2 π 1

5 sec^{2} (arccos (- 0.96414 \dots) + 2 π 1) sin (arccos (- 0.96414 \dots) + 2 π 1) - cos (arccos (- 0.96414 \dots) + 2 π 1) = 0

Refine

1.96382 \dots = 0

\Rightarrow False

Check the solution

- arccos (- 0.96414 \dots) + 2 πn :

True

- arccos (- 0.96414 \dots) + 2 πn

Plug in

n = 1

- arccos (- 0.96414 \dots) + 2 π 1

For

5 sec^{2} (θ) sin (θ) - cos (θ) = 0

plug in

θ = - arccos (- 0.96414 \dots) + 2 π 1

5 sec^{2} (- arccos (- 0.96414 \dots) + 2 π 1) sin (- arccos (- 0.96414 \dots) + 2 π 1) - cos (- arccos (- 0.96414 \dots) + 2 π 1) = 0

Refine

0 = 0

\Rightarrow True

θ = arccos (0.96414 \dots) + 2 πn, θ = - arccos (- 0.96414 \dots) + 2 πn

Show solutions in decimal form

θ = 0.19049 \dots + 2 πn, θ = - 2.95110 \dots + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 5sec^2(θ)sin(θ)-cos(θ)=0 ?
The general solution for 5sec^2(θ)sin(θ)-cos(θ)=0 is θ=0.19049…+2pin,θ=-2.95110…+2pin