What is the general solution for 5cos(x)+2sin(x)=2 ?

The general solution for 5cos(x)+2sin(x)=2 is x=-0.80978…+2pin,x= pi/2+2pin

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5cos(x)+2sin(x)=2

5 cos (x) + 2 sin (x) = 2

x = - 0.80978 \dots + 2 πn, x = \frac{π}{2} + 2 πn

Solution steps

5 cos (x) + 2 sin (x) = 2

Subtract

2 sin (x)

from both sides

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

Square both sides

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

Subtract

(2 - 2 sin (x))^{2}

from both sides

25 cos^{2} (x) - 4 + 8 sin (x) - 4 sin^{2} (x) = 0

Rewrite using trig identities

21 - 29 sin^{2} (x) + 8 sin (x) = 0

Solve by substitution

sin (x) = - \frac{21}{29}, sin (x) = 1

sin (x) = - \frac{21}{29} : x = arcsin (- \frac{21}{29}) + 2 πn, x = π + arcsin (\frac{21}{29}) + 2 πn

sin (x) = 1 : x = \frac{π}{2} + 2 πn

Combine all the solutions

x = arcsin (- \frac{21}{29}) + 2 πn, x = π + arcsin (\frac{21}{29}) + 2 πn, x = \frac{π}{2} + 2 πn

Verify solutions by plugging them into the original equation

x = arcsin (- \frac{21}{29}) + 2 πn, x = \frac{π}{2} + 2 πn

Show solutions in decimal form

x = - 0.80978 \dots + 2 πn, x = \frac{π}{2} + 2 πn

2 sin (2 x) = 0

2 cos (x) + cos (2 x) = 0

x, - sin (x + y) = 0

cot (θ) = - \frac{1}{2}

sin (2 θ) + sin (θ) + 2 cos (θ) + 1 = 0, 0 \leq θ \leq 2 π

What is the general solution for 5cos(x)+2sin(x)=2 ?
The general solution for 5cos(x)+2sin(x)=2 is x=-0.80978…+2pin,x= pi/2+2pin