What is the general solution for cos(x+40)=0.3 ?

The general solution for cos(x+40)=0.3 is x=1.26610…+360n-40,x=360-1.26610…+360n-40

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cos(x+40)=0.3

cos (x + 4 0^{\circ}) = 0.3

x = 1.26610 \dots + 36 0^{\circ} n - 4 0^{\circ}, x = 36 0^{\circ} - 1.26610 \dots + 36 0^{\circ} n - 4 0^{\circ}

Radians

x = 1.26610 \dots - \frac{2 π}{9} + 2 πn, x = 2 π - 1.26610 \dots - \frac{2 π}{9} + 2 πn

Solution steps

cos (x + 4 0^{\circ}) = 0.3

Apply trig inverse properties

cos (x + 4 0^{\circ}) = 0.3

General solutions for

cos (x + 4 0^{\circ}) = 0.3

cos (x) = a \Rightarrow x = arccos (a) + 36 0^{\circ} n, x = 36 0^{\circ} - arccos (a) + 36 0^{\circ} n

x + 4 0^{\circ} = arccos (0.3) + 36 0^{\circ} n, x + 4 0^{\circ} = 36 0^{\circ} - arccos (0.3) + 36 0^{\circ} n

x + 4 0^{\circ} = arccos (0.3) + 36 0^{\circ} n, x + 4 0^{\circ} = 36 0^{\circ} - arccos (0.3) + 36 0^{\circ} n

Solve

x + 4 0^{\circ} = arccos (0.3) + 36 0^{\circ} n : x = arccos (0.3) + 36 0^{\circ} n - 4 0^{\circ}

x + 4 0^{\circ} = arccos (0.3) + 36 0^{\circ} n

Move

4 0^{\circ}

to the right side

x + 4 0^{\circ} = arccos (0.3) + 36 0^{\circ} n

Subtract

4 0^{\circ}

from both sides

x + 4 0^{\circ} - 4 0^{\circ} = arccos (0.3) + 36 0^{\circ} n - 4 0^{\circ}

Simplify

x = arccos (0.3) + 36 0^{\circ} n - 4 0^{\circ}

x = arccos (0.3) + 36 0^{\circ} n - 4 0^{\circ}