What is the general solution for 6sin(4x)+4=5,0<= x<= 2pi ?

The general solution for 6sin(4x)+4=5,0<= x<= 2pi is

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6sin(4x)+4=5,0<= x<= 2pi

6 sin (4 x) + 4 = 5, 0 \leq x \leq 2 π

x = \frac{0.16744 \dots}{4}, x = \frac{π - 0.16744 \dots}{4}, x = \frac{0.16744 \dots + 2 π}{4}, x = \frac{3 π - 0.16744 \dots}{4}, x = \frac{0.16744 \dots + 4 π}{4}, x = \frac{5 π - 0.16744 \dots}{4}, x = \frac{0.16744 \dots + 6 π}{4}, x = \frac{7 π - 0.16744 \dots}{4}

Solution steps

6 sin (4 x) + 4 = 5, 0 \leq x \leq 2 π

Move

4

to the right side

6 sin (4 x) = 1

Divide both sides by

6

sin (4 x) = \frac{1}{6}

Apply trig inverse properties

4 x = arcsin (\frac{1}{6}) + 2 πn, 4 x = π - arcsin (\frac{1}{6}) + 2 πn

Solve

4 x = arcsin (\frac{1}{6}) + 2 πn : x = \frac{arcsin ( \frac{1}{6} )}{4} + \frac{πn}{2}

Solve

4 x = π - arcsin (\frac{1}{6}) + 2 πn : x = \frac{π}{4} - \frac{arcsin ( \frac{1}{6} )}{4} + \frac{πn}{2}

x = \frac{arcsin ( \frac{1}{6} )}{4} + \frac{πn}{2}, x = \frac{π}{4} - \frac{arcsin ( \frac{1}{6} )}{4} + \frac{πn}{2}

Solutions for the range

0 \leq x \leq 2 π

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

Show solutions in decimal form

x = \frac{0.16744 \dots}{4}, x = \frac{π - 0.16744 \dots}{4}, x = \frac{0.16744 \dots + 2 π}{4}, x = \frac{3 π - 0.16744 \dots}{4}, x = \frac{0.16744 \dots + 4 π}{4}, x = \frac{5 π - 0.16744 \dots}{4}, x = \frac{0.16744 \dots + 6 π}{4}, x = \frac{7 π - 0.16744 \dots}{4}

3 tan (x) = 3

t, y = sin (t)

sec^{2} (q) = sec (q) + 2

sin (θ) + 4 = 3

cos (3 x) = \frac{1}{3}

What is the general solution for 6sin(4x)+4=5,0<= x<= 2pi ?
The general solution for 6sin(4x)+4=5,0<= x<= 2pi is