What is the general solution for 3sin(x)tan(x)=8 ?

The general solution for 3sin(x)tan(x)=8 is x=1.23095…+2pin,x=2pi-1.23095…+2pin

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3sin(x)tan(x)=8

3 sin (x) tan (x) = 8

x = 1.23095 \dots + 2 πn, x = 2 π - 1.23095 \dots + 2 πn

Solution steps

3 sin (x) tan (x) = 8

Subtract

8

from both sides

3 sin (x) tan (x) - 8 = 0

Express with sin, cos

3 sin (x) \frac{sin ( x )}{cos ( x )} - 8 = 0

Simplify

3 sin (x) \frac{sin ( x )}{cos ( x )} - 8 : \frac{3 sin ^{2} ( x ) - 8 cos ( x )}{cos ( x )}

\frac{3 sin ^{2} ( x ) - 8 cos ( x )}{cos ( x )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

3 sin^{2} (x) - 8 cos (x) = 0

Add

8 cos (x)

to both sides

3 sin^{2} (x) = 8 cos (x)

Square both sides

(3 sin^{2} (x))^{2} = (8 cos (x))^{2}

Subtract

(8 cos (x))^{2}

from both sides

9 sin^{4} (x) - 64 cos^{2} (x) = 0

Factor

9 sin^{4} (x) - 64 cos^{2} (x) : (3 sin^{2} (x) + 8 cos (x)) (3 sin^{2} (x) - 8 cos (x))

(3 sin^{2} (x) + 8 cos (x)) (3 sin^{2} (x) - 8 cos (x)) = 0

Solving each part separately

3 sin^{2} (x) + 8 cos (x) = 0 or 3 sin^{2} (x) - 8 cos (x) = 0

3 sin^{2} (x) + 8 cos (x) = 0 : x = arccos (- \frac{1}{3}) + 2 πn, x = - arccos (- \frac{1}{3}) + 2 πn

3 sin^{2} (x) - 8 cos (x) = 0 : x = arccos (\frac{1}{3}) + 2 πn, x = 2 π - arccos (\frac{1}{3}) + 2 πn

Combine all the solutions

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

Verify solutions by plugging them into the original equation

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

Show solutions in decimal form

x = 1.23095 \dots + 2 πn, x = 2 π - 1.23095 \dots + 2 πn

tan^{2} (x) - 4 = 0

2 sin^{2} (θ) + sin (θ) - 1 = 0, 0 \leq θ < 2 π

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

cot (θ) = \frac{1}{3}

- 3 sin^{2} (θ) + 4 sin (θ) - 7 = - 6

What is the general solution for 3sin(x)tan(x)=8 ?
The general solution for 3sin(x)tan(x)=8 is x=1.23095…+2pin,x=2pi-1.23095…+2pin