15=-40sin(18(x-10))

Solution

15 = - 40 sin (18 (x - 10))

x = - \frac{0.38439 \dots}{18} + \frac{πn}{9} + 10, x = \frac{π + 0.38439 \dots}{18} + \frac{πn}{9} + 10

Solution steps

15 = - 40 sin (18 (x - 10))

Switch sides

- 40 sin (18 (x - 10)) = 15

Divide both sides by

- 40

sin (18 (x - 10)) = - \frac{3}{8}

Apply trig inverse properties

18 (x - 10) = arcsin (- \frac{3}{8}) + 2 πn, 18 (x - 10) = π + arcsin (\frac{3}{8}) + 2 πn

Solve

18 (x - 10) = arcsin (- \frac{3}{8}) + 2 πn : x = - \frac{arcsin ( \frac{3}{8} )}{18} + \frac{πn}{9} + 10

Solve

18 (x - 10) = π + arcsin (\frac{3}{8}) + 2 πn : x = \frac{π + arcsin ( \frac{3}{8} )}{18} + \frac{πn}{9} + 10

x = - \frac{arcsin ( \frac{3}{8} )}{18} + \frac{πn}{9} + 10, x = \frac{π + arcsin ( \frac{3}{8} )}{18} + \frac{πn}{9} + 10

Show solutions in decimal form

x = - \frac{0.38439 \dots}{18} + \frac{πn}{9} + 10, x = \frac{π + 0.38439 \dots}{18} + \frac{πn}{9} + 10

tan^{2} (x) - tan (x) - 6 = 0

\frac{1}{sec ( 2 x ) - 1} - \frac{1}{sec ( 2 x ) + 1} = 6

sin (2 π + x) - sin (2 π - x) = - 1

θ, ma = T sin (θ) - m g

sin (x) = \frac{16}{20}