What is the general solution for cos(x-15^0)=((sqrt(2)))/2 ?

The general solution for cos(x-15^0)=((sqrt(2)))/2 is x= pi/4+2pin+1,x=(7pi)/4+2pin+1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

cos(x-15^0)=((sqrt(2)))/2

cos (x - 1 5^{0}) = \frac{( 2 )}{2}

x = \frac{π}{4} + 2 πn + 1, x = \frac{7 π}{4} + 2 πn + 1

Degrees

x = 102.29577 \dots^{\circ} + 36 0^{\circ} n, x = 372.29577 \dots^{\circ} + 36 0^{\circ} n

Solution steps

cos (x - 1 5^{0}) = \frac{( 2 )}{2}

General solutions for

cos (x - 1 5^{0}) = \frac{2}{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 - 1 5^{0} = \frac{π}{4} + 2 πn, x - 1 5^{0} = \frac{7 π}{4} + 2 πn

x - 1 5^{0} = \frac{π}{4} + 2 πn, x - 1 5^{0} = \frac{7 π}{4} + 2 πn

Solve

x - 1 5^{0} = \frac{π}{4} + 2 πn : x = \frac{π}{4} + 2 πn + 1

x - 1 5^{0} = \frac{π}{4} + 2 πn

Apply rule

a^{0} = 1, a \neq = 0

1 5^{0} = 1

x - 1 = \frac{π}{4} + 2 πn

Move

1

to the right side

x - 1 = \frac{π}{4} + 2 πn

Add

1

to both sides

x - 1 + 1 = \frac{π}{4} + 2 πn + 1

Simplify

x = \frac{π}{4} + 2 πn + 1

x = \frac{π}{4} + 2 πn + 1