What is the general solution for sin(x)-sqrt(3)cos(x)=1,0<= x<2pi ?

Q: What is the general solution for sin(x)-sqrt(3)cos(x)=1,0<= x<2pi ?

The general solution for sin(x)-sqrt(3)cos(x)=1,0<= x<2pi is x=(7pi)/6 ,x= pi/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sin(x)-sqrt(3)cos(x)=1,0<= x<2pi

Solution

sin (x) - 3 cos (x) = 1, 0 \leq x < 2 π

Solution

x = \frac{7 π}{6}, x = \frac{π}{2}

Degrees

x = 21 0^{\circ}, x = 9 0^{\circ}

Solution steps

sin (x) - 3 cos (x) = 1, 0 \leq x < 2 π

Add

3 cos (x)

to both sides

sin (x) = 1 + 3 cos (x)

Square both sides

sin^{2} (x) = (1 + 3 cos (x))^{2}

Subtract

(1 + 3 cos (x))^{2}

from both sides

sin^{2} (x) - 1 - 23 cos (x) - 3 cos^{2} (x) = 0

Rewrite using trig identities

- 1 + sin^{2} (x) - 3 cos^{2} (x) - 2 cos (x) 3

Use the Pythagorean identity:

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

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

= - 3 cos^{2} (x) - 23 cos (x) - cos^{2} (x)

Simplify

= - 4 cos^{2} (x) - 23 cos (x)

- 4 cos^{2} (x) - 2 cos (x) 3 = 0

Solve by substitution

- 4 cos^{2} (x) - 2 cos (x) 3 = 0

Let:

cos (x) = u

- 4 u^{2} - 2 u 3 = 0

- 4 u^{2} - 2 u 3 = 0 : u = - \frac{3}{2}, u = 0

- 4 u^{2} - 2 u 3 = 0

Write in the standard form

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

- 4 u^{2} - 23 u = 0

Solve with the quadratic formula

- 4 u^{2} - 23 u = 0

Quadratic Equation Formula:

For

a = - 4, b = - 23, c = 0

u_{1, 2} = \frac{- ( - 2 3 ) \pm ( - 2 3 ) ^{2} - 4 ( - 4 ) \cdot 0}{2 ( - 4 )}

u_{1, 2} = \frac{- ( - 2 3 ) \pm ( - 2 3 ) ^{2} - 4 ( - 4 ) \cdot 0}{2 ( - 4 )}

(- 23)^{2} - 4 (- 4) \cdot 0 = 23

(- 23)^{2} - 4 (- 4) \cdot 0

Apply rule

- (- a) = a

= (- 23)^{2} + 4 \cdot 4 \cdot 0

(- 23)^{2} = 2^{2} \cdot 3

(- 23)^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 23)^{2} = (23)^{2}

= (23)^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

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

(3)^{2} : 3

Apply radical rule:

a = a^{\frac{1}{2}}

= (3^{\frac{1}{2}})^{2}

Apply exponent rule:

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

= 3^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

Multiply fractions:

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

= \frac{1 \cdot 2}{2}

Cancel the common factor:

2

= 1

= 3

= 2^{2} \cdot 3

4 \cdot 4 \cdot 0 = 0

4 \cdot 4 \cdot 0

Apply rule

0 \cdot a = 0

= 0

= 2^{2} \cdot 3 + 0

2^{2} \cdot 3 + 0 = 2^{2} \cdot 3

= 2^{2} \cdot 3

Apply radical rule:

assuming

a \geq 0, b \geq 0

= 3 2^{2}

Apply radical rule:

assuming

a \geq 0

2^{2} = 2

= 23

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

Separate the solutions

u_{1} = \frac{- ( - 2 3 ) + 2 3}{2 ( - 4 )}, u_{2} = \frac{- ( - 2 3 ) - 2 3}{2 ( - 4 )}

u = \frac{- ( - 2 3 ) + 2 3}{2 ( - 4 )} : - \frac{3}{2}

\frac{- ( - 2 3 ) + 2 3}{2 ( - 4 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{2 3 + 2 3}{- 2 \cdot 4}

Add similar elements:

23 + 23 = 43

= \frac{4 3}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{4 3}{- 8}

Apply the fraction rule:

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

= - \frac{4 3}{8}

Cancel the common factor:

4

= - \frac{3}{2}

u = \frac{- ( - 2 3 ) - 2 3}{2 ( - 4 )} : 0

\frac{- ( - 2 3 ) - 2 3}{2 ( - 4 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{2 3 - 2 3}{- 2 \cdot 4}

Add similar elements:

23 - 23 = 0

= \frac{0}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{0}{- 8}

Apply the fraction rule:

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

= - \frac{0}{8}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= - 0

= 0

The solutions to the quadratic equation are:

u = - \frac{3}{2}, u = 0

Substitute back

u = cos (x)

cos (x) = - \frac{3}{2}, cos (x) = 0

cos (x) = - \frac{3}{2}, cos (x) = 0

cos (x) = - \frac{3}{2}, 0 \leq x < 2 π : x = \frac{5 π}{6}, x = \frac{7 π}{6}

cos (x) = - \frac{3}{2}, 0 \leq x < 2 π

General solutions for

cos (x) = - \frac{3}{2}

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 = \frac{5 π}{6} + 2 πn, x = \frac{7 π}{6} + 2 πn

x = \frac{5 π}{6} + 2 πn, x = \frac{7 π}{6} + 2 πn

Solutions for the range

0 \leq x < 2 π

x = \frac{5 π}{6}, x = \frac{7 π}{6}

cos (x) = 0, 0 \leq x < 2 π : x = \frac{π}{2}, x = \frac{3 π}{2}

cos (x) = 0, 0 \leq x < 2 π

General solutions for

cos (x) = 0

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 = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Solutions for the range

0 \leq x < 2 π

x = \frac{π}{2}, x = \frac{3 π}{2}

Combine all the solutions

x = \frac{5 π}{6}, x = \frac{7 π}{6}, x = \frac{π}{2}, x = \frac{3 π}{2}

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

sin (x) - 3 cos (x) = 1

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

Check the solution

\frac{5 π}{6} :

False

\frac{5 π}{6}

Plug in

n = 1

\frac{5 π}{6}

For

sin (x) - 3 cos (x) = 1

plug in

x = \frac{5 π}{6}

sin (\frac{5 π}{6}) - 3 cos (\frac{5 π}{6}) = 1

Refine

2 = 1

\Rightarrow False

Check the solution

\frac{7 π}{6} :

True

\frac{7 π}{6}

Plug in

n = 1

\frac{7 π}{6}

For

sin (x) - 3 cos (x) = 1

plug in

x = \frac{7 π}{6}

sin (\frac{7 π}{6}) - 3 cos (\frac{7 π}{6}) = 1

Refine

1 = 1

\Rightarrow True

Check the solution

\frac{π}{2} :

True

\frac{π}{2}

Plug in

n = 1

\frac{π}{2}

For

sin (x) - 3 cos (x) = 1

plug in

x = \frac{π}{2}

sin (\frac{π}{2}) - 3 cos (\frac{π}{2}) = 1

Refine

1 = 1

\Rightarrow True

Check the solution

\frac{3 π}{2} :

False

\frac{3 π}{2}

Plug in

n = 1

\frac{3 π}{2}

For

sin (x) - 3 cos (x) = 1

plug in

x = \frac{3 π}{2}

sin (\frac{3 π}{2}) - 3 cos (\frac{3 π}{2}) = 1

Refine

- 1 = 1

\Rightarrow False

x = \frac{7 π}{6}, x = \frac{π}{2}

Popular Examples

sin(α)+1=cos(α)

sin (α) + 1 = cos (α)

(sin(115))/(53)=(sin(S))/(83)

\frac{sin ( 11 5 ^{\circ} )}{53} = \frac{sin ( S )}{83}

2cos(x)-tan(x)=0

2 cos (x) - tan (x) = 0

0=3sin(x)

0 = 3 sin (x)

sin(θ/2)=0.4

sin (\frac{θ}{2}) = 0.4

Frequently Asked Questions (FAQ)

What is the general solution for sin(x)-sqrt(3)cos(x)=1,0<= x<2pi ?
The general solution for sin(x)-sqrt(3)cos(x)=1,0<= x<2pi is x=(7pi)/6 ,x= pi/2