What is the general solution for 1=lsin(x),2=lcos(x) ?

The general solution for 1=lsin(x),2=lcos(x) is

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1=lsin(x),2=lcos(x)

1 = l sin (x), 2 = l cos (x)

x = arcsin (\frac{1}{l}), x = arcsin (- \frac{1}{l}) + π, x = arcsin (\frac{1}{l}) + 2 π, x = 3 π + arcsin (- \frac{1}{l}), x = arcsin (\frac{1}{l}) + 4 π, x = 5 π + arcsin (- \frac{1}{l}), x = arcsin (\frac{1}{l}) + 6 π, x = 7 π + arcsin (- \frac{1}{l}), x = arcsin (\frac{1}{l}) + 8 π, x = 9 π + arcsin (- \frac{1}{l}), x = arcsin (\frac{1}{l}) - 2 π, x = - π + arcsin (- \frac{1}{l}), x = arcsin (\frac{1}{l}) - 4 π, x = - 3 π + arcsin (- \frac{1}{l}), x = arcsin (\frac{1}{l}) - 6 π, x = - 5 π + arcsin (- \frac{1}{l}), x = arcsin (\frac{1}{l}) - 8 π, x = - 7 π + arcsin (- \frac{1}{l})

Solution steps

1 = l sin (x), 2 = l cos (x)

Switch sides

l sin (x) = 1

Divide both sides by

l; l \neq = 0

l sin (x) = 1

Divide both sides by

l; l \neq = 0

\frac{l sin ( x )}{l} = \frac{1}{l}; l \neq = 0

Simplify

sin (x) = \frac{1}{l}; l \neq = 0

sin (x) = \frac{1}{l}; l \neq = 0

Apply trig inverse properties

sin (x) = \frac{1}{l}

General solutions for

sin (x) = \frac{1}{l}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (\frac{1}{l}) + 2 πn, x = π + arcsin (- \frac{1}{l}) + 2 πn

x = arcsin (\frac{1}{l}) + 2 πn, x = π + arcsin (- \frac{1}{l}) + 2 πn

Solutions for the range

2 = l cos (x)

x = arcsin (\frac{1}{l}), x = arcsin (- \frac{1}{l}) + π, x = arcsin (\frac{1}{l}) + 2 π, x = 3 π + arcsin (- \frac{1}{l}), x = arcsin (\frac{1}{l}) + 4 π, x = 5 π + arcsin (- \frac{1}{l}), x = arcsin (\frac{1}{l}) + 6 π, x = 7 π + arcsin (- \frac{1}{l}), x = arcsin (\frac{1}{l}) + 8 π, x = 9 π + arcsin (- \frac{1}{l}), x = arcsin (\frac{1}{l}) - 2 π, x = - π + arcsin (- \frac{1}{l}), x = arcsin (\frac{1}{l}) - 4 π, x = - 3 π + arcsin (- \frac{1}{l}), x = arcsin (\frac{1}{l}) - 6 π, x = - 5 π + arcsin (- \frac{1}{l}), x = arcsin (\frac{1}{l}) - 8 π, x = - 7 π + arcsin (- \frac{1}{l})

y, t = 2 cot (6 0^{\circ} y + 3 0^{\circ})

sec (y) = - 2

- 1 + 2 sin^{2} (x) = 0

θ, y = - 2 + cos (3 θ)

cos (2 x) + 5 \cdot cos (x) = - 3

What is the general solution for 1=lsin(x),2=lcos(x) ?
The general solution for 1=lsin(x),2=lcos(x) is