What is the general solution for 4sin(t/2)-1=1,-pi/2 <= t<= pi/2 ?

Q: What is the general solution for 4sin(t/2)-1=1,-pi/2 <= t<= pi/2 ?

The general solution for 4sin(t/2)-1=1,-pi/2 <= t<= pi/2 is t= pi/3

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4sin(t/2)-1=1,-pi/2 <= t<= pi/2

Solution

4 sin (\frac{t}{2}) - 1 = 1, - \frac{π}{2} \leq t \leq \frac{π}{2}

Solution

t = \frac{π}{3}

Degrees

t = 6 0^{\circ}

Solution steps

4 sin (\frac{t}{2}) - 1 = 1, - \frac{π}{2} \leq t \leq \frac{π}{2}

Move

1

to the right side

4 sin (\frac{t}{2}) - 1 = 1

Add

1

to both sides

4 sin (\frac{t}{2}) - 1 + 1 = 1 + 1

Simplify

4 sin (\frac{t}{2}) = 2

4 sin (\frac{t}{2}) = 2

Divide both sides by

4

4 sin (\frac{t}{2}) = 2

Divide both sides by

4

\frac{4 sin ( \frac{t}{2} )}{4} = \frac{2}{4}

Simplify

sin (\frac{t}{2}) = \frac{1}{2}

sin (\frac{t}{2}) = \frac{1}{2}

General solutions for

sin (\frac{t}{2}) = \frac{1}{2}

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

\frac{t}{2} = \frac{π}{6} + 2 πn, \frac{t}{2} = \frac{5 π}{6} + 2 πn

\frac{t}{2} = \frac{π}{6} + 2 πn, \frac{t}{2} = \frac{5 π}{6} + 2 πn

Solve

\frac{t}{2} = \frac{π}{6} + 2 πn : t = \frac{π}{3} + 4 πn

\frac{t}{2} = \frac{π}{6} + 2 πn

Multiply both sides by

2

\frac{t}{2} = \frac{π}{6} + 2 πn

Multiply both sides by

2

\frac{2 t}{2} = 2 \cdot \frac{π}{6} + 2 \cdot 2 πn

Simplify

\frac{2 t}{2} = 2 \cdot \frac{π}{6} + 2 \cdot 2 πn

Simplify

\frac{2 t}{2} : t

\frac{2 t}{2}

Divide the numbers:

\frac{2}{2} = 1

= t

Simplify

2 \cdot \frac{π}{6} + 2 \cdot 2 πn : \frac{π}{3} + 4 πn

2 \cdot \frac{π}{6} + 2 \cdot 2 πn

2 \cdot \frac{π}{6} = \frac{π}{3}

2 \cdot \frac{π}{6}

Multiply fractions:

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

= \frac{π 2}{6}

Cancel the common factor:

2

= \frac{π}{3}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

Multiply the numbers:

2 \cdot 2 = 4

= 4 πn

= \frac{π}{3} + 4 πn

t = \frac{π}{3} + 4 πn

t = \frac{π}{3} + 4 πn

t = \frac{π}{3} + 4 πn

Solve

\frac{t}{2} = \frac{5 π}{6} + 2 πn : t = \frac{5 π}{3} + 4 πn

\frac{t}{2} = \frac{5 π}{6} + 2 πn

Multiply both sides by

2

\frac{t}{2} = \frac{5 π}{6} + 2 πn

Multiply both sides by

2

\frac{2 t}{2} = 2 \cdot \frac{5 π}{6} + 2 \cdot 2 πn

Simplify

\frac{2 t}{2} = 2 \cdot \frac{5 π}{6} + 2 \cdot 2 πn

Simplify

\frac{2 t}{2} : t

\frac{2 t}{2}

Divide the numbers:

\frac{2}{2} = 1

= t

Simplify

2 \cdot \frac{5 π}{6} + 2 \cdot 2 πn : \frac{5 π}{3} + 4 πn

2 \cdot \frac{5 π}{6} + 2 \cdot 2 πn

2 \cdot \frac{5 π}{6} = \frac{5 π}{3}

2 \cdot \frac{5 π}{6}

Multiply fractions:

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

= \frac{5 π 2}{6}

Multiply the numbers:

5 \cdot 2 = 10

= \frac{10 π}{6}

Cancel the common factor:

2

= \frac{5 π}{3}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

Multiply the numbers:

2 \cdot 2 = 4

= 4 πn

= \frac{5 π}{3} + 4 πn

t = \frac{5 π}{3} + 4 πn

t = \frac{5 π}{3} + 4 πn

t = \frac{5 π}{3} + 4 πn

t = \frac{π}{3} + 4 πn, t = \frac{5 π}{3} + 4 πn

Solutions for the range

- \frac{π}{2} \leq t \leq \frac{π}{2}

t = \frac{π}{3}

Popular Examples

-6cos(θ)-10=-7

- 6 cos (θ) - 10 = - 7

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

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

tanh(x)=1

tanh (x) = 1

tan(θ)=sqrt(1/3)

tan (θ) = \frac{1}{3}

sin(θ)= 20/29

sin (θ) = \frac{20}{29}

Frequently Asked Questions (FAQ)

What is the general solution for 4sin(t/2)-1=1,-pi/2 <= t<= pi/2 ?
The general solution for 4sin(t/2)-1=1,-pi/2 <= t<= pi/2 is t= pi/3