What is the general solution for 4sin(x)=cos(x)-2 ?

The general solution for 4sin(x)=cos(x)-2 is x=-2.39016…+2pin,x=2pi-0.26146…+2pin

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4sin(x)=cos(x)-2

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

x = - 2.39016 \dots + 2 πn, x = 2 π - 0.26146 \dots + 2 πn

Solution steps

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

Square both sides

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

Subtract

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

from both sides

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

Rewrite using trig identities

12 - 17 cos^{2} (x) + 4 cos (x) = 0

Solve by substitution

cos (x) = - \frac{2 ( 2 13 - 1 )}{17}, cos (x) = \frac{2 ( 1 + 2 13 )}{17}

cos (x) = - \frac{2 ( 2 13 - 1 )}{17} : x = arccos (- \frac{2 ( 2 13 - 1 )}{17}) + 2 πn, x = - arccos (- \frac{2 ( 2 13 - 1 )}{17}) + 2 πn

cos (x) = \frac{2 ( 1 + 2 13 )}{17} : x = arccos (\frac{2 ( 1 + 2 13 )}{17}) + 2 πn, x = 2 π - arccos (\frac{2 ( 1 + 2 13 )}{17}) + 2 πn

Combine all the solutions

x = arccos (- \frac{2 ( 2 13 - 1 )}{17}) + 2 πn, x = - arccos (- \frac{2 ( 2 13 - 1 )}{17}) + 2 πn, x = arccos (\frac{2 ( 1 + 2 13 )}{17}) + 2 πn, x = 2 π - arccos (\frac{2 ( 1 + 2 13 )}{17}) + 2 πn

Verify solutions by plugging them into the original equation

x = - arccos (- \frac{2 ( 2 13 - 1 )}{17}) + 2 πn, x = 2 π - arccos (\frac{2 ( 1 + 2 13 )}{17}) + 2 πn

Show solutions in decimal form

x = - 2.39016 \dots + 2 πn, x = 2 π - 0.26146 \dots + 2 πn

sin (θ) = 1 - cos (θ)

cos (θ) - 2 sin (θ) cos (θ) = 0

x, cos (x y) = sin (x + y)

cot (x) = \frac{3}{4}

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

What is the general solution for 4sin(x)=cos(x)-2 ?
The general solution for 4sin(x)=cos(x)-2 is x=-2.39016…+2pin,x=2pi-0.26146…+2pin