What is the general solution for 3sin(4x)+5=6,0<= x<= pi/2 ?

The general solution for 3sin(4x)+5=6,0<= x<= pi/2 is x=(0.33983…)/4 ,x=(pi-0.33983…)/4

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3sin(4x)+5=6,0<= x<= pi/2

3 sin (4 x) + 5 = 6, 0 \leq x \leq \frac{π}{2}

x = \frac{0.33983 \dots}{4}, x = \frac{π - 0.33983 \dots}{4}

Solution steps

3 sin (4 x) + 5 = 6, 0 \leq x \leq \frac{π}{2}

Move

5

to the right side

3 sin (4 x) = 1

Divide both sides by

3

sin (4 x) = \frac{1}{3}

Apply trig inverse properties

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

Solve

4 x = arcsin (\frac{1}{3}) + 2 πn : x = \frac{arcsin ( \frac{1}{3} )}{4} + \frac{πn}{2}

Solve

4 x = π - arcsin (\frac{1}{3}) + 2 πn : x = \frac{π}{4} - \frac{arcsin ( \frac{1}{3} )}{4} + \frac{πn}{2}

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

Solutions for the range

0 \leq x \leq \frac{π}{2}

x = \frac{arcsin ( \frac{1}{3} )}{4}, x = \frac{π - arcsin ( \frac{1}{3} )}{4}

Show solutions in decimal form

x = \frac{0.33983 \dots}{4}, x = \frac{π - 0.33983 \dots}{4}

cos (z + 1) = 0

4 csc^{2} (\frac{x}{2}) - 6 csc (\frac{x}{2}) - 4 = 0

2 tan (θ) + sec^{2} (θ) = 4

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

csc (x) = 7

What is the general solution for 3sin(4x)+5=6,0<= x<= pi/2 ?
The general solution for 3sin(4x)+5=6,0<= x<= pi/2 is x=(0.33983…)/4 ,x=(pi-0.33983…)/4