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

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

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Popular >

sec(x)+cos(x)=sin(x)tan(x)

Solution

sec (x) + cos (x) = sin (x) tan (x)

Solution

No Solution for x \in R

Solution steps

sec (x) + cos (x) = sin (x) tan (x)

Subtract

sin (x) tan (x)

from both sides

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

Express with sin, cos

cos (x) + sec (x) - sin (x) tan (x)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Multiply fractions:

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

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

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

sin (x) sin (x)

Apply exponent rule:

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

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

= sin^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= sin^{2} (x)

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

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

Combine the fractions

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

Apply rule

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

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

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

cos (x) cos (x)

Apply exponent rule:

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

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

= cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= cos^{2} (x)

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

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

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

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

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

1 + cos^{2} (x) - sin^{2} (x) = 0

Rewrite using trig identities

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

Use the Double Angle identity:

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

= 1 + cos (2 x)

1 + cos (2 x) = 0

Move

1

to the right side

1 + cos (2 x) = 0

Subtract

1

from both sides

1 + cos (2 x) - 1 = 0 - 1

Simplify

cos (2 x) = - 1

cos (2 x) = - 1

General solutions for

cos (2 x) = - 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 x = π + 2 πn

2 x = π + 2 πn

Solve

2 x = π + 2 πn : x = \frac{π}{2} + πn

2 x = π + 2 πn

Divide both sides by

2

2 x = π + 2 πn

Divide both sides by

2

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

Simplify

x = \frac{π}{2} + πn

x = \frac{π}{2} + πn

x = \frac{π}{2} + πn

Since the equation is undefined for:

\frac{π}{2} + πn

No Solution for x \in R

Graph

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solvefor x,sin(x+30)=tan(10)solve for

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Frequently Asked Questions (FAQ)

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