What is the general solution for cos(x)cos(35)=1-sin(x)sin(35) ?

The general solution for cos(x)cos(35)=1-sin(x)sin(35) is x=-360n+35

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Popular Trigonometry >

cos(x)cos(35)=1-sin(x)sin(35)

Solution

cos (x) cos (3 5^{\circ}) = 1 - sin (x) sin (3 5^{\circ})

Solution

x = - 36 0^{\circ} n + 3 5^{\circ}

Radians

x = \frac{7 π}{36} - 2 πn

Solution steps

cos (x) cos (3 5^{\circ}) = 1 - sin (x) sin (3 5^{\circ})

Subtract

1 - sin (x) sin (3 5^{\circ})

from both sides

cos (3 5^{\circ}) cos (x) - 1 + sin (3 5^{\circ}) sin (x) = 0

Rewrite using trig identities

cos (3 5^{\circ}) cos (x) - 1 + sin (3 5^{\circ}) sin (x)

Use the Angle Difference identity:

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

= - 1 + cos (3 5^{\circ} - x)

- 1 + cos (3 5^{\circ} - x) = 0

Move

1

to the right side

- 1 + cos (3 5^{\circ} - x) = 0

Add

1

to both sides

- 1 + cos (3 5^{\circ} - x) + 1 = 0 + 1

Simplify

cos (3 5^{\circ} - x) = 1

cos (3 5^{\circ} - x) = 1

General solutions for

cos (3 5^{\circ} - x) = 1

cos (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

3 5^{\circ} - x = 0 + 36 0^{\circ} n

3 5^{\circ} - x = 0 + 36 0^{\circ} n

Solve

3 5^{\circ} - x = 0 + 36 0^{\circ} n : x = - 36 0^{\circ} n + 3 5^{\circ}

3 5^{\circ} - x = 0 + 36 0^{\circ} n

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

3 5^{\circ} - x = 36 0^{\circ} n

Move

3 5^{\circ}

to the right side

3 5^{\circ} - x = 36 0^{\circ} n

Subtract

3 5^{\circ}

from both sides

3 5^{\circ} - x - 3 5^{\circ} = 36 0^{\circ} n - 3 5^{\circ}

Simplify

- x = 36 0^{\circ} n - 3 5^{\circ}

- x = 36 0^{\circ} n - 3 5^{\circ}

Divide both sides by

- 1

- x = 36 0^{\circ} n - 3 5^{\circ}

Divide both sides by

- 1

\frac{- x}{- 1} = \frac{36 0 ^{\circ} n}{- 1} - \frac{3 5 ^{\circ}}{- 1}

Simplify

\frac{- x}{- 1} = \frac{36 0 ^{\circ} n}{- 1} - \frac{3 5 ^{\circ}}{- 1}

Simplify

\frac{- x}{- 1} : x

\frac{- x}{- 1}

Apply the fraction rule:

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

= \frac{x}{1}

Apply rule

\frac{a}{1} = a

= x

Simplify

\frac{36 0 ^{\circ} n}{- 1} - \frac{3 5 ^{\circ}}{- 1} : - 36 0^{\circ} n + 3 5^{\circ}

\frac{36 0 ^{\circ} n}{- 1} - \frac{3 5 ^{\circ}}{- 1}

\frac{36 0 ^{\circ} n}{- 1} = - 36 0^{\circ} n

\frac{36 0 ^{\circ} n}{- 1}

Apply the fraction rule:

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

= - \frac{36 0 ^{\circ} n}{1}

Apply rule

\frac{a}{1} = a

= - 36 0^{\circ} n

= - 36 0^{\circ} n - \frac{3 5 ^{\circ}}{- 1}

\frac{3 5 ^{\circ}}{- 1} = - 3 5^{\circ}

\frac{3 5 ^{\circ}}{- 1}

Apply the fraction rule:

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

= - \frac{3 5 ^{\circ}}{1}

Apply the fraction rule:

\frac{a}{1} = a

\frac{3 5 ^{\circ}}{1} = 3 5^{\circ}

= - 3 5^{\circ}

= - 36 0^{\circ} n - (- 3 5^{\circ})

Apply rule

- (- a) = a

= - 36 0^{\circ} n + 3 5^{\circ}

x = - 36 0^{\circ} n + 3 5^{\circ}

x = - 36 0^{\circ} n + 3 5^{\circ}

x = - 36 0^{\circ} n + 3 5^{\circ}

x = - 36 0^{\circ} n + 3 5^{\circ}

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for cos(x)cos(35)=1-sin(x)sin(35) ?
The general solution for cos(x)cos(35)=1-sin(x)sin(35) is x=-360n+35