What is the general solution for 7tan(x)=2sqrt(3)+tan(x),0<= θ<= 90 ?

The general solution for 7tan(x)=2sqrt(3)+tan(x),0<= θ<= 90 is x=30

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7tan(x)=2sqrt(3)+tan(x),0<= θ<= 90

7 tan (x) = 23 + tan (x), 0^{\circ} \leq θ \leq 9 0^{\circ}

x = 3 0^{\circ}

Radians

x = \frac{π}{6}

Solution steps

7 tan (x) = 23 + tan (x), 0^{\circ} \leq θ \leq 9 0^{\circ}

Solve by substitution

7 tan (x) = 23 + tan (x)

Let:

tan (x) = u

7 u = 23 + u

7 u = 23 + u : u = \frac{3}{3}

7 u = 23 + u

Move

u

to the left side

7 u = 23 + u

Subtract

u

from both sides

7 u - u = 23 + u - u

Simplify

6 u = 23

6 u = 23

Divide both sides by

6

6 u = 23

Divide both sides by

6

\frac{6 u}{6} = \frac{2 3}{6}

Simplify

u = \frac{3}{3}

u = \frac{3}{3}

Substitute back

u = tan (x)

tan (x) = \frac{3}{3}

tan (x) = \frac{3}{3}

tan (x) = \frac{3}{3}, 0 \leq θ \leq 9 0^{\circ} : x = 3 0^{\circ}

tan (x) = \frac{3}{3}, 0 \leq θ \leq 9 0^{\circ}

General solutions for

tan (x) = \frac{3}{3}

tan (x)

periodicity table with

18 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = 3 0^{\circ} + 18 0^{\circ} n

x = 3 0^{\circ} + 18 0^{\circ} n

Solutions for the range

0 \leq θ \leq 9 0^{\circ}

x = 3 0^{\circ}

Combine all the solutions

x = 3 0^{\circ}

sin (X) = 1

x, \frac{17}{sin ( x )} = \frac{23}{sin ( 7 5 ^{\circ} )}

3 - 2 cos^{2} (x) = 4 sin^{2} (x)

4 - 4 cos (2 x) - 3 sin (x) = 0

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

What is the general solution for 7tan(x)=2sqrt(3)+tan(x),0<= θ<= 90 ?
The general solution for 7tan(x)=2sqrt(3)+tan(x),0<= θ<= 90 is x=30