What is the general solution for 7sin(x)=cos(x)-3 ?

The general solution for 7sin(x)=cos(x)-3 is x=-2.56154…+2pin,x=2pi-0.29625…+2pin

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7sin(x)=cos(x)-3

7 sin (x) = cos (x) - 3

x = - 2.56154 \dots + 2 πn, x = 2 π - 0.29625 \dots + 2 πn

Solution steps

7 sin (x) = cos (x) - 3

Square both sides

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

Subtract

(cos (x) - 3)^{2}

from both sides

49 sin^{2} (x) - cos^{2} (x) + 6 cos (x) - 9 = 0

Rewrite using trig identities

40 - 50 cos^{2} (x) + 6 cos (x) = 0

Solve by substitution

cos (x) = - \frac{- 3 + 7 41}{50}, cos (x) = \frac{3 + 7 41}{50}

cos (x) = - \frac{- 3 + 7 41}{50} : x = arccos (- \frac{- 3 + 7 41}{50}) + 2 πn, x = - arccos (- \frac{- 3 + 7 41}{50}) + 2 πn

cos (x) = \frac{3 + 7 41}{50} : x = arccos (\frac{3 + 7 41}{50}) + 2 πn, x = 2 π - arccos (\frac{3 + 7 41}{50}) + 2 πn

Combine all the solutions

x = arccos (- \frac{- 3 + 7 41}{50}) + 2 πn, x = - arccos (- \frac{- 3 + 7 41}{50}) + 2 πn, x = arccos (\frac{3 + 7 41}{50}) + 2 πn, x = 2 π - arccos (\frac{3 + 7 41}{50}) + 2 πn

Verify solutions by plugging them into the original equation

x = - arccos (- \frac{- 3 + 7 41}{50}) + 2 πn, x = 2 π - arccos (\frac{3 + 7 41}{50}) + 2 πn

Show solutions in decimal form

x = - 2.56154 \dots + 2 πn, x = 2 π - 0.29625 \dots + 2 πn

sin (a) = \frac{3}{4}

sin (7 x) + sin (x) = 0, 0 \leq x \leq 2 π

t, - 2 sin (t) + 2 cos (2 t) = 0

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

cos^{2} (x) = tan (x)

What is the general solution for 7sin(x)=cos(x)-3 ?
The general solution for 7sin(x)=cos(x)-3 is x=-2.56154…+2pin,x=2pi-0.29625…+2pin