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

The general solution for sqrt(2)sin(x)=sqrt(3)-cos(x),0<= x<= 2pi is x=0.95531…

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Popular >

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

Solution

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

Solution

x = 0.95531 \dots

Degrees

x = 54.73561 \dots^{\circ}

Solution steps

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

Square both sides

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

Subtract

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

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

- 3 - cos^{2} (x) + 2 (1 - cos^{2} (x)) + 2 cos (x) 3 : 23 cos (x) - 3 cos^{2} (x) - 1

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

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

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

Multiply the numbers:

2 \cdot 1 = 2

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

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

Simplify

- 3 - cos^{2} (x) + 2 - 2 cos^{2} (x) + 2 cos (x) 3 : 23 cos (x) - 3 cos^{2} (x) - 1

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

Group like terms

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

Add similar elements:

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

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

Add/Subtract the numbers:

- 3 + 2 = - 1

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

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

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

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

Solve by substitution

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

Let:

cos (x) = u

- 1 - 3 u^{2} + 2 u 3 = 0

- 1 - 3 u^{2} + 2 u 3 = 0 : u = \frac{3}{3}

- 1 - 3 u^{2} + 2 u 3 = 0

Write in the standard form

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

- 3 u^{2} + 23 u - 1 = 0

Solve with the quadratic formula

- 3 u^{2} + 23 u - 1 = 0

Quadratic Equation Formula:

For

a = - 3, b = 23, c = - 1

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

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

(23)^{2} - 4 (- 3) (- 1) = 0

(23)^{2} - 4 (- 3) (- 1)

Apply rule

- (- a) = a

= (23)^{2} - 4 \cdot 3 \cdot 1

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

(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 3 \cdot 1 = 12

4 \cdot 3 \cdot 1

Multiply the numbers:

4 \cdot 3 \cdot 1 = 12

= 12

= 2^{2} \cdot 3 - 12

2^{2} \cdot 3 = 12

2^{2} \cdot 3

2^{2} = 4

= 4 \cdot 3

Multiply the numbers:

4 \cdot 3 = 12

= 12

= 12 - 12

Subtract the numbers:

12 - 12 = 0

= 0

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

u = \frac{- 2 3}{2 ( - 3 )}

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

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

Remove parentheses:

(- a) = - a

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

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 2 3}{- 6}

Apply the fraction rule:

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

= \frac{2 3}{6}

Cancel the common factor:

2

= \frac{3}{3}

u = \frac{3}{3}

The solution to the quadratic equation is:

u = \frac{3}{3}

Substitute back

u = cos (x)

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

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

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

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

Apply trig inverse properties

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

General solutions for

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

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos (\frac{3}{3}) + 2 πn, x = 2 π - arccos (\frac{3}{3}) + 2 πn

x = arccos (\frac{3}{3}) + 2 πn, x = 2 π - arccos (\frac{3}{3}) + 2 πn

Solutions for the range

0 \leq x \leq 2 π

x = arccos (\frac{3}{3}), x = 2 π - arccos (\frac{3}{3})

Combine all the solutions

x = arccos (\frac{3}{3}), x = 2 π - arccos (\frac{3}{3})

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

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

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

Check the solution

arccos (\frac{3}{3}) :

True

arccos (\frac{3}{3})

Plug in

n = 1

arccos (\frac{3}{3})

For

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

plug in

x = arccos (\frac{3}{3})

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

Refine

1.15470 \dots = 1.15470 \dots

\Rightarrow True

Check the solution

2 π - arccos (\frac{3}{3}) :

False

2 π - arccos (\frac{3}{3})

Plug in

n = 1

2 π - arccos (\frac{3}{3})

For

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

plug in

x = 2 π - arccos (\frac{3}{3})

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

Refine

- 1.15470 \dots = 1.15470 \dots

\Rightarrow False

x = arccos (\frac{3}{3})

Show solutions in decimal form

x = 0.95531 \dots

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Frequently Asked Questions (FAQ)

What is the general solution for sqrt(2)sin(x)=sqrt(3)-cos(x),0<= x<= 2pi ?
The general solution for sqrt(2)sin(x)=sqrt(3)-cos(x),0<= x<= 2pi is x=0.95531…