What is the general solution for sqrt(3)sin(x)+cos(x)=2 ?

The general solution for sqrt(3)sin(x)+cos(x)=2 is x= pi/3+2pin

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sqrt(3)sin(x)+cos(x)=2

Solution

3 sin (x) + cos (x) = 2

Solution

x = \frac{π}{3} + 2 πn

Degrees

x = 6 0^{\circ} + 36 0^{\circ} n

Solution steps

3 sin (x) + cos (x) = 2

Subtract

cos (x)

from both sides

3 sin (x) = 2 - cos (x)

Square both sides

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

Subtract

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

from both sides

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

Rewrite using trig identities

- 4 - cos^{2} (x) + 3 sin^{2} (x) + 4 cos (x)

Use the Pythagorean identity:

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

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

= - 4 - cos^{2} (x) + 3 (1 - cos^{2} (x)) + 4 cos (x)

Simplify

- 4 - cos^{2} (x) + 3 (1 - cos^{2} (x)) + 4 cos (x) : 4 cos (x) - 4 cos^{2} (x) - 1

- 4 - cos^{2} (x) + 3 (1 - cos^{2} (x)) + 4 cos (x)

Expand

3 (1 - cos^{2} (x)) : 3 - 3 cos^{2} (x)

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

Apply the distributive law:

a (b - c) = ab - a c

a = 3, b = 1, c = cos^{2} (x)

= 3 \cdot 1 - 3 cos^{2} (x)

Multiply the numbers:

3 \cdot 1 = 3

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

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

Simplify

- 4 - cos^{2} (x) + 3 - 3 cos^{2} (x) + 4 cos (x) : 4 cos (x) - 4 cos^{2} (x) - 1

- 4 - cos^{2} (x) + 3 - 3 cos^{2} (x) + 4 cos (x)

Group like terms

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

Add similar elements:

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

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

Add/Subtract the numbers:

- 4 + 3 = - 1

= 4 cos (x) - 4 cos^{2} (x) - 1

= 4 cos (x) - 4 cos^{2} (x) - 1

= 4 cos (x) - 4 cos^{2} (x) - 1

- 1 + 4 cos (x) - 4 cos^{2} (x) = 0

Solve by substitution

- 1 + 4 cos (x) - 4 cos^{2} (x) = 0

Let:

cos (x) = u

- 1 + 4 u - 4 u^{2} = 0

- 1 + 4 u - 4 u^{2} = 0 : u = \frac{1}{2}

- 1 + 4 u - 4 u^{2} = 0

Write in the standard form

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

- 4 u^{2} + 4 u - 1 = 0

Solve with the quadratic formula

- 4 u^{2} + 4 u - 1 = 0

Quadratic Equation Formula:

For

a = - 4, b = 4, c = - 1

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

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

4^{2} - 4 (- 4) (- 1) = 0

4^{2} - 4 (- 4) (- 1)

Apply rule

- (- a) = a

= 4^{2} - 4 \cdot 4 \cdot 1

Multiply the numbers:

4 \cdot 4 \cdot 1 = 16

= 4^{2} - 16

4^{2} = 16

= 16 - 16

Subtract the numbers:

16 - 16 = 0

= 0

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

u = \frac{- 4}{2 ( - 4 )}

\frac{- 4}{2 ( - 4 )} = \frac{1}{2}

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

Remove parentheses:

(- a) = - a

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 4}{- 8}

Apply the fraction rule:

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

= \frac{4}{8}

Cancel the common factor:

4

= \frac{1}{2}

u = \frac{1}{2}

The solution to the quadratic equation is:

u = \frac{1}{2}

Substitute back

u = cos (x)

cos (x) = \frac{1}{2}

cos (x) = \frac{1}{2}

cos (x) = \frac{1}{2} : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

cos (x) = \frac{1}{2}

General solutions for

cos (x) = \frac{1}{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{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Combine all the solutions

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

3 sin (x) + cos (x) = 2

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

Check the solution

\frac{π}{3} + 2 πn :

True

\frac{π}{3} + 2 πn

Plug in

n = 1

\frac{π}{3} + 2 π 1

For

3 sin (x) + cos (x) = 2

plug in

x = \frac{π}{3} + 2 π 1

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

Refine

2 = 2

\Rightarrow True

Check the solution

\frac{5 π}{3} + 2 πn :

False

\frac{5 π}{3} + 2 πn

Plug in

n = 1

\frac{5 π}{3} + 2 π 1

For

3 sin (x) + cos (x) = 2

plug in

x = \frac{5 π}{3} + 2 π 1

3 sin (\frac{5 π}{3} + 2 π 1) + cos (\frac{5 π}{3} + 2 π 1) = 2

Refine

- 1 = 2

\Rightarrow False

x = \frac{π}{3} + 2 πn

Popular Examples

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

-3sin^2(θ)+3=-cos(θ)+2

- 3 sin^{2} (θ) + 3 = - cos (θ) + 2

tan^2(x)-4sec(x)+5=0

tan^{2} (x) - 4 sec (x) + 5 = 0

cot(θ)= 3/4

cot (θ) = \frac{3}{4}

sin(x)cos(x)=-1/2

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

Frequently Asked Questions (FAQ)

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