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

The general solution for 3cos(x)-2sin(x)=1 is x=pi+1.26382…+2pin,x=0.70175…+2pin

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

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

x = π + 1.26382 \dots + 2 πn, x = 0.70175 \dots + 2 πn

Solution steps

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

Add

2 sin (x)

to both sides

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

Square both sides

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

Subtract

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

from both sides

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

Rewrite using trig identities

8 - 13 sin^{2} (x) - 4 sin (x) = 0

Solve by substitution

sin (x) = - \frac{2 ( 1 + 3 3 )}{13}, sin (x) = \frac{2 ( 3 3 - 1 )}{13}

sin (x) = - \frac{2 ( 1 + 3 3 )}{13} : x = arcsin (- \frac{2 ( 1 + 3 3 )}{13}) + 2 πn, x = π + arcsin (\frac{2 ( 1 + 3 3 )}{13}) + 2 πn

sin (x) = \frac{2 ( 3 3 - 1 )}{13} : x = arcsin (\frac{2 ( 3 3 - 1 )}{13}) + 2 πn, x = π - arcsin (\frac{2 ( 3 3 - 1 )}{13}) + 2 πn

Combine all the solutions

x = arcsin (- \frac{2 ( 1 + 3 3 )}{13}) + 2 πn, x = π + arcsin (\frac{2 ( 1 + 3 3 )}{13}) + 2 πn, x = arcsin (\frac{2 ( 3 3 - 1 )}{13}) + 2 πn, x = π - arcsin (\frac{2 ( 3 3 - 1 )}{13}) + 2 πn

Verify solutions by plugging them into the original equation

x = π + arcsin (\frac{2 ( 1 + 3 3 )}{13}) + 2 πn, x = arcsin (\frac{2 ( 3 3 - 1 )}{13}) + 2 πn

Show solutions in decimal form

x = π + 1.26382 \dots + 2 πn, x = 0.70175 \dots + 2 πn

\frac{sin ( 9 5 ^{\circ} )}{27} = \frac{sin ( b )}{16}

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

sinh (x) = \frac{12}{35}

sin^{2} (x) + \frac{1}{sec ( x )} = \frac{5}{4}

2 cos (\frac{1}{5} θ) = 3

What is the general solution for 3cos(x)-2sin(x)=1 ?
The general solution for 3cos(x)-2sin(x)=1 is x=pi+1.26382…+2pin,x=0.70175…+2pin