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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2sin^2(x)+5sin(x)-3=0,0<= x<= 2pi

2 sin^{2} (x) + 5 sin (x) - 3 = 0, 0 \leq x \leq 2 π

x = \frac{π}{6}, x = \frac{5 π}{6}

Solution steps

2 sin^{2} (x) + 5 sin (x) - 3 = 0, 0 \leq x \leq 2 π

Solve by substitution

sin (x) = \frac{1}{2}, sin (x) = - 3

sin (x) = \frac{1}{2}, 0 \leq x \leq 2 π : x = \frac{π}{6}, x = \frac{5 π}{6}

sin (x) = - 3, 0 \leq x \leq 2 π :

No Solution

Combine all the solutions

x = \frac{π}{6}, x = \frac{5 π}{6}

sin (x) = \frac{123 sin ( 7 0 ^{\circ} )}{184.5}

cot^{2} (θ) = cot (θ), 0 \leq θ \leq 2 π

cos (x) = \frac{9}{19}

cos (α) = \frac{5}{3}

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

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