2sin(x)tan(x)+1=tan(x)+2sin(x)

Solution

2 sin (x) tan (x) + 1 = tan (x) + 2 sin (x)

x = \frac{π}{4} + πn, x = 2 π - 2 πn - \frac{7 π}{6}, x = 2 π - 2 πn - \frac{11 π}{6}

Solution steps

2 sin (x) tan (x) + 1 = tan (x) + 2 sin (x)

Subtract

tan (x) + 2 sin (x)

from both sides

2 sin (x) tan (x) + 1 - tan (x) - 2 sin (x) = 0

Rewrite using trig identities

1 - tan (x) + 2 sin (- x + 2 π) - 2 sin (- x + 2 π) tan (x) = 0

Factor

1 - tan (x) + 2 sin (- x + 2 π) - 2 sin (- x + 2 π) tan (x) : - (tan (x) - 1) (2 sin (- x + 2 π) + 1)

- (tan (x) - 1) (2 sin (- x + 2 π) + 1) = 0

Solving each part separately

tan (x) - 1 = 0 or 2 sin (- x + 2 π) + 1 = 0

tan (x) - 1 = 0 : x = \frac{π}{4} + πn

2 sin (- x + 2 π) + 1 = 0 : x = 2 π - 2 πn - \frac{7 π}{6}, x = 2 π - 2 πn - \frac{11 π}{6}

Combine all the solutions

x = \frac{π}{4} + πn, x = 2 π - 2 πn - \frac{7 π}{6}, x = 2 π - 2 πn - \frac{11 π}{6}

sec (y) = (\frac{5}{4})

\frac{sin ( 6 5 ^{\circ} )}{5} = \frac{sin ( x )}{4}

2 - 6.5 sin (x) = 0

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

1.33 \cdot sin (6 0^{\circ}) = 1.52 \cdot sin (x)