What is the general solution for sin(x)+cos(x)=tan(x)+1 ?

The general solution for sin(x)+cos(x)=tan(x)+1 is x=2pin,x=(3pi)/4+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sin(x)+cos(x)=tan(x)+1

Solution

sin (x) + cos (x) = tan (x) + 1

Solution

x = 2 πn, x = \frac{3 π}{4} + πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

sin (x) + cos (x) = tan (x) + 1

Subtract

tan (x) + 1

from both sides

sin (x) + cos (x) - tan (x) - 1 = 0

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Convert element to fraction:

1 = \frac{1 cos ( x )}{cos ( x )}, cos (x) = \frac{cos ( x ) cos ( x )}{cos ( x )}, sin (x) = \frac{sin ( x ) cos ( x )}{cos ( x )}

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

Since the denominators are equal, combine the fractions:

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

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

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

- 1 \cdot cos (x) + cos (x) cos (x) + sin (x) cos (x) - sin (x)

1 \cdot cos (x) = cos (x)

1 \cdot cos (x)

Multiply:

1 \cdot cos (x) = cos (x)

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

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

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

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

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

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

Factor

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

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

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

Factor out

sin (x)

from

sin (x) cos (x) - sin (x) : sin (x) (cos (x) - 1)

sin (x) cos (x) - sin (x)

Factor out common term

sin (x)

= sin (x) (cos (x) - 1)

Factor out

cos (x)

from

cos^{2} (x) - cos (x) : cos (x) (cos (x) - 1)

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

Apply exponent rule:

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

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

= cos (x) cos (x) - cos (x)

Factor out common term

cos (x)

= cos (x) (cos (x) - 1)

= sin (x) (cos (x) - 1) + cos (x) (cos (x) - 1)

Factor out common term

cos (x) - 1

= (cos (x) - 1) (sin (x) + cos (x))

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

Solving each part separately

cos (x) - 1 = 0 or sin (x) + cos (x) = 0

cos (x) - 1 = 0 : x = 2 πn

cos (x) - 1 = 0

Move

1

to the right side

cos (x) - 1 = 0

Add

1

to both sides

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

Simplify

cos (x) = 1

cos (x) = 1

General solutions for

cos (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}

x = 0 + 2 πn

x = 0 + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

sin (x) + cos (x) = 0 : x = \frac{3 π}{4} + πn

sin (x) + cos (x) = 0

Rewrite using trig identities

sin (x) + cos (x) = 0

Divide both sides by

cos (x), cos (x) \neq = 0

\frac{sin ( x ) + cos ( x )}{cos ( x )} = \frac{0}{cos ( x )}

Simplify

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

Use the basic trigonometric identity:

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

tan (x) + 1 = 0

tan (x) + 1 = 0

Move

1

to the right side

tan (x) + 1 = 0

Subtract

1

from both sides

tan (x) + 1 - 1 = 0 - 1

Simplify

tan (x) = - 1

tan (x) = - 1

General solutions for

tan (x) = - 1

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

Combine all the solutions

x = 2 πn, x = \frac{3 π}{4} + πn

Popular Examples

cos(x)= 1/4 ,0<= x<= 2pi

cos (x) = \frac{1}{4}, 0 \leq x \leq 2 π

tan^2(x)+1=a^2

tan^{2} (x) + 1 = a^{2}

2=2cos(t)

2 = 2 cos (t)

50sin(45)=600sin(θ)

50 sin (4 5^{\circ}) = 600 sin (θ)

solvefor a,tan(a*x^{0.5})=-x^{0.5}solve for

a, tan (a \cdot x^{0.5}) = - x^{0.5}

Frequently Asked Questions (FAQ)

What is the general solution for sin(x)+cos(x)=tan(x)+1 ?
The general solution for sin(x)+cos(x)=tan(x)+1 is x=2pin,x=(3pi)/4+pin