What is the general solution for 7cos(t)+5sin(t)=0 ?

The general solution for 7cos(t)+5sin(t)=0 is t=-0.95054…+pin

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7cos(t)+5sin(t)=0

7 cos (t) + 5 sin (t) = 0

t = - 0.95054 \dots + πn

Degrees

t = - 54.46232 \dots^{\circ} + 18 0^{\circ} n

Solution steps

7 cos (t) + 5 sin (t) = 0

Rewrite using trig identities

7 cos (t) + 5 sin (t) = 0

Divide both sides by

cos (t), cos (t) \neq = 0

\frac{7 cos ( t ) + 5 sin ( t )}{cos ( t )} = \frac{0}{cos ( t )}

Simplify

7 + \frac{5 sin ( t )}{cos ( t )} = 0

Use the basic trigonometric identity:

\frac{sin ( x )}{cos ( x )} = tan (x)

7 + 5 tan (t) = 0

7 + 5 tan (t) = 0

Move

7

to the right side

7 + 5 tan (t) = 0

Subtract

7

from both sides

7 + 5 tan (t) - 7 = 0 - 7

Simplify

5 tan (t) = - 7

5 tan (t) = - 7

Divide both sides by

5

5 tan (t) = - 7

Divide both sides by

5

\frac{5 tan ( t )}{5} = \frac{- 7}{5}

Simplify

tan (t) = - \frac{7}{5}

tan (t) = - \frac{7}{5}

Apply trig inverse properties

tan (t) = - \frac{7}{5}

General solutions for

tan (t) = - \frac{7}{5}

tan (x) = - a \Rightarrow x = arctan (- a) + πn

t = arctan (- \frac{7}{5}) + πn

t = arctan (- \frac{7}{5}) + πn

Show solutions in decimal form

t = - 0.95054 \dots + πn

cos (X) = 0.4848

\frac{π}{3} = 3 + 2 sin (2 t)

1 + sin (2 x) = 5 (sin (x) + cos (x))

x, y = I n [tan (5 x^{2} + 3 x + 2)]

cos (2 x) = \frac{3}{4} - 5 cos^{2} (x)

What is the general solution for 7cos(t)+5sin(t)=0 ?
The general solution for 7cos(t)+5sin(t)=0 is t=-0.95054…+pin