What is the general solution for sec(θ)+cos(θ)=sin(θ)tan(θ) ?

The general solution for sec(θ)+cos(θ)=sin(θ)tan(θ) is No Solution for θ\in\mathbb{R}

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Popular >

sec(θ)+cos(θ)=sin(θ)tan(θ)

Solution

sec (θ) + cos (θ) = sin (θ) tan (θ)

Solution

No Solution for θ \in R

Solution steps

sec (θ) + cos (θ) = sin (θ) tan (θ)

Subtract

sin (θ) tan (θ)

from both sides

sec (θ) + cos (θ) - sin (θ) tan (θ) = 0

Express with sin, cos

cos (θ) + sec (θ) - sin (θ) tan (θ)

Use the basic trigonometric identity:

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

= cos (θ) + \frac{1}{cos ( θ )} - sin (θ) tan (θ)

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

= cos (θ) + \frac{1}{cos ( θ )} - sin (θ) \frac{sin ( θ )}{cos ( θ )}

Simplify

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

cos (θ) + \frac{1}{cos ( θ )} - sin (θ) \frac{sin ( θ )}{cos ( θ )}

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

sin (θ) \frac{sin ( θ )}{cos ( θ )}

Multiply fractions:

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

= \frac{sin ( θ ) sin ( θ )}{cos ( θ )}

sin (θ) sin (θ) = sin^{2} (θ)

sin (θ) sin (θ)

Apply exponent rule:

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

sin (θ) sin (θ) = sin^{1 + 1} (θ)

= sin^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= sin^{2} (θ)

= \frac{sin ^{2} ( θ )}{cos ( θ )}

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

Combine the fractions

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

Apply rule

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

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

cos (θ) cos (θ) + 1 - sin^{2} (θ)

cos (θ) cos (θ) = cos^{2} (θ)

cos (θ) cos (θ)

Apply exponent rule:

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

cos (θ) cos (θ) = cos^{1 + 1} (θ)

= cos^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= cos^{2} (θ)

= cos^{2} (θ) + 1 - sin^{2} (θ)

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

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

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

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

1 + cos^{2} (θ) - sin^{2} (θ) = 0

Rewrite using trig identities

1 + cos^{2} (θ) - sin^{2} (θ)

Use the Double Angle identity:

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

= 1 + cos (2 θ)

1 + cos (2 θ) = 0

Move

1

to the right side

1 + cos (2 θ) = 0

Subtract

1

from both sides

1 + cos (2 θ) - 1 = 0 - 1

Simplify

cos (2 θ) = - 1

cos (2 θ) = - 1

General solutions for

cos (2 θ) = - 1

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}

2 θ = π + 2 πn

2 θ = π + 2 πn

Solve

2 θ = π + 2 πn : θ = \frac{π}{2} + πn

2 θ = π + 2 πn

Divide both sides by

2

2 θ = π + 2 πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{π}{2} + \frac{2 πn}{2}

Simplify

θ = \frac{π}{2} + πn

θ = \frac{π}{2} + πn

θ = \frac{π}{2} + πn

Since the equation is undefined for:

\frac{π}{2} + πn

No Solution for θ \in R

Popular Examples

cos(x)=0.98

cos (x) = 0.98

tan(3x+pi/2)=-1

tan (3 x + \frac{π}{2}) = - 1

15=60sin(pi/(15[0-d]))+75

15 = 60 sin (\frac{π}{15 [ 0 - d ]}) + 75

cos(θ)=(2sqrt(5))/(sqrt(58))

cos (θ) = \frac{2 5}{58}

tan(θ)=(m^2-m^1)/(1+m^2*m^1),m^2

tan (θ) = \frac{m ^{2} - m ^{1}}{1 + m ^{2} \cdot m ^{1}}, m^{2}

Frequently Asked Questions (FAQ)

What is the general solution for sec(θ)+cos(θ)=sin(θ)tan(θ) ?
The general solution for sec(θ)+cos(θ)=sin(θ)tan(θ) is No Solution for θ\in\mathbb{R}