What is the general solution for sin(x)+sqrt(3)cos(x)=-sqrt(2) ?

The general solution for sin(x)+sqrt(3)cos(x)=-sqrt(2) is x=2.87979…+2pin,x=-1.83259…+2pin

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sin(x)+sqrt(3)cos(x)=-sqrt(2)

sin (x) + 3 cos (x) = - 2

x = 2.87979 \dots + 2 πn, x = - 1.83259 \dots + 2 πn

Solution steps

sin (x) + 3 cos (x) = - 2

Subtract

3 cos (x)

from both sides

sin (x) = - 2 - 3 cos (x)

Square both sides

sin^{2} (x) = (- 2 - 3 cos (x))^{2}

Subtract

(- 2 - 3 cos (x))^{2}

from both sides

sin^{2} (x) - 2 - 26 cos (x) - 3 cos^{2} (x) = 0

Rewrite using trig identities

- 1 - 4 cos^{2} (x) - 2 cos (x) 6 = 0

Solve by substitution

cos (x) = - \frac{6 + 2}{4}, cos (x) = - \frac{6 - 2}{4}

cos (x) = - \frac{6 + 2}{4} : x = arccos (- \frac{6 + 2}{4}) + 2 πn, x = - arccos (- \frac{6 + 2}{4}) + 2 πn

cos (x) = - \frac{6 - 2}{4} : x = arccos (- \frac{6 - 2}{4}) + 2 πn, x = - arccos (- \frac{6 - 2}{4}) + 2 πn

Combine all the solutions

x = arccos (- \frac{6 + 2}{4}) + 2 πn, x = - arccos (- \frac{6 + 2}{4}) + 2 πn, x = arccos (- \frac{6 - 2}{4}) + 2 πn, x = - arccos (- \frac{6 - 2}{4}) + 2 πn

Verify solutions by plugging them into the original equation

x = arccos (- \frac{6 + 2}{4}) + 2 πn, x = - arccos (- \frac{6 - 2}{4}) + 2 πn

Show solutions in decimal form

x = 2.87979 \dots + 2 πn, x = - 1.83259 \dots + 2 πn

cot (θ) + 4 csc (θ) = 6

sin (x) 2 = - sin (x)

tan^{3} (x) = 1

sin^{2} (x) - 9 cos (x) - 9 = 0

sin (a) = \frac{8}{17}

What is the general solution for sin(x)+sqrt(3)cos(x)=-sqrt(2) ?
The general solution for sin(x)+sqrt(3)cos(x)=-sqrt(2) is x=2.87979…+2pin,x=-1.83259…+2pin