What is the general solution for tan(3x+25)*160-1=0 ?

The general solution for tan(3x+25)*160-1=0 is x=(180n)/3-8.33333…+(0.34387…)/3

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tan(3x+25)*160-1=0

tan (3 x + 2 5^{\circ}) \cdot 16 0^{\circ} - 1 = 0

x = \frac{18 0 ^{\circ} n}{3} - 8.33333 \dots^{\circ} + \frac{0.34387 \dots}{3}

Solution steps

tan (3 x + 2 5^{\circ}) \cdot 16 0^{\circ} - 1 = 0

Move

1

to the right side

tan (3 x + 2 5^{\circ}) 16 0^{\circ} = 1

Multiply both sides by

9

144 0^{\circ} tan (3 x + 2 5^{\circ}) = 9

Divide both sides by

144 0^{\circ}

tan (3 x + 2 5^{\circ}) = \frac{9}{144 0 ^{\circ}}

Apply trig inverse properties

3 x + 2 5^{\circ} = arctan (\frac{9}{144 0 ^{\circ}}) + 18 0^{\circ} n

Solve

3 x + 2 5^{\circ} = arctan (\frac{9}{144 0 ^{\circ}}) + 18 0^{\circ} n : x = \frac{18 0 ^{\circ} n}{3} - 8.33333 \dots^{\circ} + \frac{arctan ( \frac{9}{144 0 ^{\circ}} )}{3}

x = \frac{18 0 ^{\circ} n}{3} - 8.33333 \dots^{\circ} + \frac{arctan ( \frac{9}{144 0 ^{\circ}} )}{3}

Show solutions in decimal form

x = \frac{18 0 ^{\circ} n}{3} - 8.33333 \dots^{\circ} + \frac{0.34387 \dots}{3}

5 cos (x) = 1

cos (θ) = \frac{2}{11}

3 cos (x) + 5 = 7, 0 \leq x \leq 2 π

9 sin (\frac{x}{2}) + 9 cos (x) = 0

- 5 cos (x) + 5 sin (x) = \frac{5 2}{2}

What is the general solution for tan(3x+25)*160-1=0 ?
The general solution for tan(3x+25)*160-1=0 is x=(180n)/3-8.33333…+(0.34387…)/3