What is the general solution for 5sin(θ/2)=5cos(θ/2) ?

The general solution for 5sin(θ/2)=5cos(θ/2) is θ= pi/2+2pin

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5sin(θ/2)=5cos(θ/2)

5 sin (\frac{θ}{2}) = 5 cos (\frac{θ}{2})

θ = \frac{π}{2} + 2 πn

Degrees

θ = 9 0^{\circ} + 36 0^{\circ} n

Solution steps

5 sin (\frac{θ}{2}) = 5 cos (\frac{θ}{2})

Rewrite using trig identities

5 sin (\frac{θ}{2}) = 5 cos (\frac{θ}{2})

Divide both sides by

cos (\frac{θ}{2}), cos (\frac{θ}{2}) \neq = 0

\frac{5 sin ( \frac{θ}{2} )}{cos ( \frac{θ}{2} )} = \frac{5 cos ( \frac{θ}{2} )}{cos ( \frac{θ}{2} )}

Simplify

\frac{5 sin ( \frac{θ}{2} )}{cos ( \frac{θ}{2} )} = 5

Use the basic trigonometric identity:

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

5 tan (\frac{θ}{2}) = 5

5 tan (\frac{θ}{2}) = 5

Divide both sides by

5

5 tan (\frac{θ}{2}) = 5

Divide both sides by

5

\frac{5 tan ( \frac{θ}{2} )}{5} = \frac{5}{5}

Simplify

tan (\frac{θ}{2}) = 1

tan (\frac{θ}{2}) = 1

General solutions for

tan (\frac{θ}{2}) = 1

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

\frac{θ}{2} = \frac{π}{4} + πn

\frac{θ}{2} = \frac{π}{4} + πn

Solve

\frac{θ}{2} = \frac{π}{4} + πn : θ = \frac{π}{2} + 2 πn

\frac{θ}{2} = \frac{π}{4} + πn

Multiply both sides by

2

\frac{θ}{2} = \frac{π}{4} + πn

Multiply both sides by

2

\frac{2 θ}{2} = 2 \cdot \frac{π}{4} + 2 πn

Simplify

\frac{2 θ}{2} = 2 \cdot \frac{π}{4} + 2 πn

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

2 \cdot \frac{π}{4} + 2 πn : \frac{π}{2} + 2 πn

2 \cdot \frac{π}{4} + 2 πn

2 \cdot \frac{π}{4} = \frac{π}{2}

2 \cdot \frac{π}{4}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{π 2}{4}

Cancel the common factor:

2

= \frac{π}{2}

= \frac{π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn

tan (x) = \frac{80}{50}

- cos (x) + 4 cos (2 x) = 0

sin^{2} (x) + sin (x) - 2 = 0, 0 \leq x < 2 π

sin^{2} (x) + 5 sin (x) = 0

tan (α) = - \frac{4}{3}, \frac{π}{2} < α < π, cos (β) = \frac{1}{2}, 0 < β < \frac{π}{2}

What is the general solution for 5sin(θ/2)=5cos(θ/2) ?
The general solution for 5sin(θ/2)=5cos(θ/2) is θ= pi/2+2pin