What is the general solution for 25sin(θ)-1.5cos(θ)=20 ?

The general solution for 25sin(θ)-1.5cos(θ)=20 is θ=2.27661…+2pin,θ=0.98483…+2pin

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25sin(θ)-1.5cos(θ)=20

25 sin (θ) - 1.5 cos (θ) = 20

θ = 2.27661 \dots + 2 πn, θ = 0.98483 \dots + 2 πn

Solution steps

25 sin (θ) - 1.5 cos (θ) = 20

Add

1.5 cos (θ)

to both sides

25 sin (θ) = 20 + 1.5 cos (θ)

Square both sides

(25 sin (θ))^{2} = (20 + 1.5 cos (θ))^{2}

Subtract

(20 + 1.5 cos (θ))^{2}

from both sides

625 sin^{2} (θ) - 400 - 60 cos (θ) - 2.25 cos^{2} (θ) = 0

Rewrite using trig identities

225 - 60 cos (θ) - 627.25 cos^{2} (θ) = 0

Solve by substitution

cos (θ) = - \frac{6000 + 5681250000}{125450}, cos (θ) = \frac{5681250000 - 6000}{125450}

cos (θ) = - \frac{6000 + 5681250000}{125450} : θ = arccos (- \frac{6000 + 5681250000}{125450}) + 2 πn, θ = - arccos (- \frac{6000 + 5681250000}{125450}) + 2 πn

cos (θ) = \frac{5681250000 - 6000}{125450} : θ = arccos (\frac{5681250000 - 6000}{125450}) + 2 πn, θ = 2 π - arccos (\frac{5681250000 - 6000}{125450}) + 2 πn

Combine all the solutions

θ = arccos (- \frac{6000 + 5681250000}{125450}) + 2 πn, θ = - arccos (- \frac{6000 + 5681250000}{125450}) + 2 πn, θ = arccos (\frac{5681250000 - 6000}{125450}) + 2 πn, θ = 2 π - arccos (\frac{5681250000 - 6000}{125450}) + 2 πn

Verify solutions by plugging them into the original equation

θ = arccos (- \frac{6000 + 5681250000}{125450}) + 2 πn, θ = arccos (\frac{5681250000 - 6000}{125450}) + 2 πn

Show solutions in decimal form

θ = 2.27661 \dots + 2 πn, θ = 0.98483 \dots + 2 πn

sin (\frac{2 θ}{3}) = \frac{1}{2}, 0 \leq θ \leq 2 π

1 - sin^{2} (x) = e^{\frac{1}{2} l n (1 - c o s^{2} (x))}

cot^{2} (x) = tan (\frac{1}{2} x)

cos (C) = \frac{3}{4}

cos (x - \frac{π}{6}) - sin (x) = 0

What is the general solution for 25sin(θ)-1.5cos(θ)=20 ?
The general solution for 25sin(θ)-1.5cos(θ)=20 is θ=2.27661…+2pin,θ=0.98483…+2pin