What is the general solution for (-(3*cos(Q^1)))/2 =0 ?

The general solution for (-(3*cos(Q^1)))/2 =0 is Q= pi/2+2pin,Q=(3pi)/2+2pin

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(-(3*cos(Q^1)))/2 =0

\frac{- ( 3 \cdot cos ( Q ^{1} ) )}{2} = 0

Q = \frac{π}{2} + 2 πn, Q = \frac{3 π}{2} + 2 πn

Degrees

Q = 9 0^{\circ} + 36 0^{\circ} n, Q = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

\frac{- ( 3 cos ( Q ^{1} ) )}{2} = 0

Multiply both sides by

2

\frac{- 3 cos ( Q ^{1} )}{2} = 0

Multiply both sides by

2

\frac{2 ( - 3 cos ( Q ^{1} ) )}{2} = 0 \cdot 2

Simplify

- 3 cos (Q^{1}) = 0

- 3 cos (Q^{1}) = 0

Divide both sides by

- 3

- 3 cos (Q^{1}) = 0

Divide both sides by

- 3

\frac{- 3 cos ( Q ^{1} )}{- 3} = \frac{0}{- 3}

Simplify

cos (Q^{1}) = 0

cos (Q^{1}) = 0

General solutions for

cos (Q^{1}) = 0

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

Q^{1} = \frac{π}{2} + 2 πn, Q^{1} = \frac{3 π}{2} + 2 πn

Q^{1} = \frac{π}{2} + 2 πn, Q^{1} = \frac{3 π}{2} + 2 πn

Solve

Q^{1} = \frac{π}{2} + 2 πn : Q = \frac{π}{2} + 2 πn

Q^{1} = \frac{π}{2} + 2 πn

Apply rule

a^{1} = a

Q = \frac{π}{2} + 2 πn

Solve

Q^{1} = \frac{3 π}{2} + 2 πn : Q = \frac{3 π}{2} + 2 πn

Q^{1} = \frac{3 π}{2} + 2 πn

Apply rule

a^{1} = a

Q = \frac{3 π}{2} + 2 πn

Q = \frac{π}{2} + 2 πn, Q = \frac{3 π}{2} + 2 πn

tan^{2} (x) + 3 tan (x) = 0, 0^{\circ} \leq x < 36 0^{\circ}

\frac{1}{2} (6.50941) + cos (θ) (10.40478) = 12

- 8 = - 4 + 4 tan (θ)

cos (x) = 0.2588

2 tan^{2} (x) + 3 tan (x) + 1 = 0

What is the general solution for (-(3*cos(Q^1)))/2 =0 ?
The general solution for (-(3*cos(Q^1)))/2 =0 is Q= pi/2+2pin,Q=(3pi)/2+2pin