What is the general solution for 1+sin(2x)=5(sin(x)+cos(x)) ?

The general solution for 1+sin(2x)=5(sin(x)+cos(x)) is x=(3pi)/4+pin

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Popular >

1+sin(2x)=5(sin(x)+cos(x))

Solution

1 + sin (2 x) = 5 (sin (x) + cos (x))

Solution

x = \frac{3 π}{4} + πn

Degrees

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

Solution steps

1 + sin (2 x) = 5 (sin (x) + cos (x))

Subtract

5 (sin (x) + cos (x))

from both sides

1 + sin (2 x) - 5 (sin (x) + cos (x)) = 0

Rewrite using trig identities

1 + sin (2 x) - (cos (x) + sin (x)) \cdot 5

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

1 + sin (2 x)

Use the Pythagorean identity:

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

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

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

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

Apply Perfect Square Formula:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

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

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

= (sin (x) + cos (x))^{2} - 5 (cos (x) + sin (x))

(cos (x) + sin (x))^{2} - (cos (x) + sin (x)) \cdot 5 = 0

Factor

(cos (x) + sin (x))^{2} - (cos (x) + sin (x)) \cdot 5 : (cos (x) + sin (x)) (cos (x) + sin (x) - 5)

(cos (x) + sin (x))^{2} - (cos (x) + sin (x)) \cdot 5

Factor out common term

(cos (x) + sin (x)) : (cos (x) + sin (x)) ((cos (x) + sin (x)) - 5)

(cos (x) + sin (x))^{2} - 5 (cos (x) + sin (x))

Apply exponent rule:

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

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

= (cos (x) + sin (x)) (cos (x) + sin (x)) - 5 (cos (x) + sin (x))

Factor out common term

(cos (x) + sin (x))

= (cos (x) + sin (x)) ((cos (x) + sin (x)) - 5)

= (cos (x) + sin (x)) ((cos (x) + sin (x)) - 5)

Refine

= (cos (x) + sin (x)) (cos (x) + sin (x) - 5)

(cos (x) + sin (x)) (cos (x) + sin (x) - 5) = 0

Solving each part separately

cos (x) + sin (x) = 0 or cos (x) + sin (x) - 5 = 0

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

cos (x) + sin (x) = 0

Rewrite using trig identities

cos (x) + sin (x) = 0

Divide both sides by

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

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

Simplify

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

Use the basic trigonometric identity:

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

1 + tan (x) = 0

1 + tan (x) = 0

Move

1

to the right side

1 + tan (x) = 0

Subtract

1

from both sides

1 + tan (x) - 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

cos (x) + sin (x) - 5 = 0 :

No Solution

cos (x) + sin (x) - 5 = 0

Rewrite using trig identities

cos (x) + sin (x) - 5

sin (x) + cos (x) = 2 sin (x + \frac{π}{4})

sin (x) + cos (x)

Rewrite as

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

Use the following trivial identity:

cos (\frac{π}{4}) = \frac{1}{2}

Use the following trivial identity:

sin (\frac{π}{4}) = \frac{1}{2}

= 2 (cos (\frac{π}{4}) sin (x) + sin (\frac{π}{4}) cos (x))

Use the Angle Sum identity:

sin (s + t) = sin (s) cos (t) + cos (s) sin (t)

= 2 sin (x + \frac{π}{4})

= - 5 + 2 sin (x + \frac{π}{4})

- 5 + 2 sin (x + \frac{π}{4}) = 0

Move

5

to the right side

- 5 + 2 sin (x + \frac{π}{4}) = 0

Add

5

to both sides

- 5 + 2 sin (x + \frac{π}{4}) + 5 = 0 + 5

Simplify

2 sin (x + \frac{π}{4}) = 5

2 sin (x + \frac{π}{4}) = 5

Divide both sides by

2

2 sin (x + \frac{π}{4}) = 5

Divide both sides by

2

\frac{2 sin ( x + \frac{π}{4} )}{2} = \frac{5}{2}

Simplify

\frac{2 sin ( x + \frac{π}{4} )}{2} = \frac{5}{2}

Simplify

\frac{2 sin ( x + \frac{π}{4} )}{2} : sin (x + \frac{π}{4})

\frac{2 sin ( x + \frac{π}{4} )}{2}

Cancel the common factor:

2

= sin (x + \frac{π}{4})

Simplify

\frac{5}{2} : \frac{5 2}{2}

\frac{5}{2}

Multiply by the conjugate

\frac{2}{2}

= \frac{5 2}{2 2}

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= \frac{5 2}{2}

sin (x + \frac{π}{4}) = \frac{5 2}{2}

sin (x + \frac{π}{4}) = \frac{5 2}{2}

sin (x + \frac{π}{4}) = \frac{5 2}{2}

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = \frac{3 π}{4} + πn

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Frequently Asked Questions (FAQ)

What is the general solution for 1+sin(2x)=5(sin(x)+cos(x)) ?
The general solution for 1+sin(2x)=5(sin(x)+cos(x)) is x=(3pi)/4+pin