What is the general solution for 3cos(x)+1=2sin(x) ?

The general solution for 3cos(x)+1=2sin(x) is x=1.26382…+2pin,x=-2.43983…+2pin

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3cos(x)+1=2sin(x)

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

x = 1.26382 \dots + 2 πn, x = - 2.43983 \dots + 2 πn

Solution steps

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

Square both sides

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

Subtract

(2 sin (x))^{2}

from both sides

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

Rewrite using trig identities

- 3 + 13 cos^{2} (x) + 6 cos (x) = 0

Solve by substitution

cos (x) = \frac{- 3 + 4 3}{13}, cos (x) = - \frac{3 + 4 3}{13}

cos (x) = \frac{- 3 + 4 3}{13} : x = arccos (\frac{- 3 + 4 3}{13}) + 2 πn, x = 2 π - arccos (\frac{- 3 + 4 3}{13}) + 2 πn

cos (x) = - \frac{3 + 4 3}{13} : x = arccos (- \frac{3 + 4 3}{13}) + 2 πn, x = - arccos (- \frac{3 + 4 3}{13}) + 2 πn

Combine all the solutions

x = arccos (\frac{- 3 + 4 3}{13}) + 2 πn, x = 2 π - arccos (\frac{- 3 + 4 3}{13}) + 2 πn, x = arccos (- \frac{3 + 4 3}{13}) + 2 πn, x = - arccos (- \frac{3 + 4 3}{13}) + 2 πn

Verify solutions by plugging them into the original equation

x = arccos (\frac{- 3 + 4 3}{13}) + 2 πn, x = - arccos (- \frac{3 + 4 3}{13}) + 2 πn

Show solutions in decimal form

x = 1.26382 \dots + 2 πn, x = - 2.43983 \dots + 2 πn

\frac{π}{6} = cos (2 x)

\frac{5}{csc ( x )} - 53 cos (x) = 0

tan (2 y) = 1

(- 6 cos (x) + 2 sin (x))^{2} + 32 sin^{2} (x) = 48

tan (x) + 5 = \frac{15 - 3}{3}

What is the general solution for 3cos(x)+1=2sin(x) ?
The general solution for 3cos(x)+1=2sin(x) is x=1.26382…+2pin,x=-2.43983…+2pin