What is the general solution for sin(x)-2cos(x)=1 ?

The general solution for sin(x)-2cos(x)=1 is x=-2.49809…+2pin,x= pi/2+2pin

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sin(x)-2cos(x)=1

sin (x) - 2 cos (x) = 1

x = - 2.49809 \dots + 2 πn, x = \frac{π}{2} + 2 πn

Solution steps

sin (x) - 2 cos (x) = 1

Add

2 cos (x)

to both sides

sin (x) = 1 + 2 cos (x)

Square both sides

sin^{2} (x) = (1 + 2 cos (x))^{2}

Subtract

(1 + 2 cos (x))^{2}

from both sides

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

Rewrite using trig identities

- 4 cos (x) - 5 cos^{2} (x) = 0

Solve by substitution

cos (x) = - \frac{4}{5}, cos (x) = 0

cos (x) = - \frac{4}{5} : x = arccos (- \frac{4}{5}) + 2 πn, x = - arccos (- \frac{4}{5}) + 2 πn

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Combine all the solutions

x = arccos (- \frac{4}{5}) + 2 πn, x = - arccos (- \frac{4}{5}) + 2 πn, x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Verify solutions by plugging them into the original equation

x = - arccos (- \frac{4}{5}) + 2 πn, x = \frac{π}{2} + 2 πn

Show solutions in decimal form

x = - 2.49809 \dots + 2 πn, x = \frac{π}{2} + 2 πn

4 cos (x) - 1 = 0

sin (2 θ) = - \frac{1}{2}, 0 \leq θ < 2 π

sin (θ) = 1 - sin (θ)

2 sin (x) - 3 = 0, 0 \leq x \leq 2 π

cos (θ) = cot (θ)

What is the general solution for sin(x)-2cos(x)=1 ?
The general solution for sin(x)-2cos(x)=1 is x=-2.49809…+2pin,x= pi/2+2pin