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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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}

Solution steps

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

Apply trig inverse properties

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}

Solve

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

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

Show solutions in decimal form

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}

cos (3 x) = - 1, 0 \leq x \leq 2 π

sin (x) = - 3 cos (x)

4 sin^{2} (x) + 2 cos^{2} (x) = 3, 0 \leq x \leq 2 π

sin (x) = - 5

sin (x) = 10

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