Solve 8cost=12sint

Question

Solve the equation

t = 2 arctan (- \frac{3 + 13}{2}) + kπ, k \in Z

Alternative Form

t \approx - 146.30993 2^{\circ} + 18 0^{\circ} k, k \in Z

Alternative Form

t \approx - 2.55359 + kπ, k \in Z

Evaluate

8 cos (t) = 12 sin (t)

Move the expression to the left side

8 cos (t) - 12 sin (t) = 0

Simplify the equation using the Weierstrass substitution

8 \times \frac{1 - tan ^{2} ( \frac{1}{2} t )}{1 + tan ^{2} ( \frac{1}{2} t )} - 12 \times \frac{2 tan ( \frac{1}{2} t )}{1 + tan ^{2} ( \frac{1}{2} t )} = 0

Solve using substitution

8 \times \frac{1 - t ^{2}}{1 + t ^{2}} - 12 \times \frac{2 t}{1 + t ^{2}} = 0

Rewrite the expression

\frac{8 ( 1 - t ^{2} )}{1 + t ^{2}} - 12 \times \frac{2 t}{1 + t ^{2}} = 0

Rewrite the expression

More Steps

\frac{8 ( 1 - t ^{2} )}{1 + t ^{2}} - \frac{24 t}{1 + t ^{2}} = 0

Multiply both sides of the equation by LCD

(\frac{8 ( 1 - t ^{2} )}{1 + t ^{2}} - \frac{24 t}{1 + t ^{2}}) (1 + t^{2}) = 0 \times (1 + t^{2})

Simplify the equation

More Steps

8 - 8 t^{2} - 24 t = 0 \times (1 + t^{2})

Any expression multiplied by 0 equals 0

8 - 8 t^{2} - 24 t = 0

Rewrite in standard form

- 8 t^{2} - 24 t + 8 = 0

Multiply both sides

8 t^{2} + 24 t - 8 = 0

Substitute a= 8,b= 24 and c= - 8 into the quadratic formula t = \frac{- b \pm b ^{2} - 4 a c}{2 a}

t = \frac{- 24 \pm 2 4 ^{2} - 4 \times 8 ( - 8 )}{2 \times 8}

Simplify the expression

t = \frac{- 24 \pm 2 4 ^{2} - 4 \times 8 ( - 8 )}{16}

Simplify the expression

More Steps

t = \frac{- 24 \pm 832}{16}

Simplify the radical expression

More Steps

t = \frac{- 24 \pm 8 13}{16}

Separate the equation into 2 possible cases

t = \frac{- 24 + 8 13}{16} t = \frac{- 24 - 8 13}{16}

Simplify the expression

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t = \frac{- 3 + 13}{2} t = \frac{- 24 - 8 13}{16}

Simplify the expression

More Steps

t = \frac{- 3 + 13}{2} t = - \frac{3 + 13}{2}

Substitute back

tan (\frac{1}{2} t) = \frac{- 3 + 13}{2} tan (\frac{1}{2} t) = - \frac{3 + 13}{2}

Calculate

More Steps

t = 2 arctan (\frac{- 3 + 13}{2}) + 2 kπ, k \in Z tan (\frac{1}{2} t) = - \frac{3 + 13}{2}

Calculate

More Steps

t = 2 arctan (\frac{- 3 + 13}{2}) + 2 kπ, k \in Z t = 2 arctan (- \frac{3 + 13}{2}) + 2 kπ, k \in Z

Check if x = π + 2 kπ, k \in Z is a solution

8 cos (π + 2 kπ) = 12 sin (π + 2 kπ)

Calculate

8 cos (π) = 12 sin (π)

Simplify

More Steps

- 8 = 12 sin (π)

Simplify

More Steps

- 8 = 0

Check the equality

false

Since x = π + 2 kπ, k \in Z is not a solution,don’t include it

t = 2 arctan (\frac{- 3 + 13}{2}) + 2 kπ, k \in Z t = 2 arctan (- \frac{3 + 13}{2}) + 2 kπ, k \in Z

Solution

t = 2 arctan (- \frac{3 + 13}{2}) + kπ, k \in Z

Alternative Form

t \approx - 146.30993 2^{\circ} + 18 0^{\circ} k, k \in Z

Alternative Form

t \approx - 2.55359 + kπ, k \in Z

Show Solution