en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4sin^2(x)+2sin(x)=2,0<= x<= 2pi

Solution

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

Solution

x = \frac{π}{6}, x = \frac{5 π}{6}, x = \frac{3 π}{2}

+1

Solution steps

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

Solve by substitution

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

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

sin (x) = - 1, 0 \leq x \leq 2 π : x = \frac{3 π}{2}

Combine all the solutions

x = \frac{π}{6}, x = \frac{5 π}{6}, x = \frac{3 π}{2}

Popular Examples

3 tan (\frac{1}{3} x) = 8

1.65^2=0.65^2+1.77^2-2(0.65)(1.77)cos(A)

1.6 5^{2} = 0.6 5^{2} + 1.7 7^{2} - 2 (0.65) (1.77) cos (A)

solvefor x,cos(3x+pi/4)= 1/(sqrt(2))solve for

x, cos (3 x + \frac{π}{4}) = \frac{1}{2}

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

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

solvefor x,cos^2(3x+pi/4)= 3/4 solve for

x, cos^{2} (3 x + \frac{π}{4}) = \frac{3}{4}

Frequently Asked Questions (FAQ)

What is the general solution for 4sin^2(x)+2sin(x)=2,0<= x<= 2pi ?
The general solution for 4sin^2(x)+2sin(x)=2,0<= x<= 2pi is x= pi/6 ,x=(5pi)/6 ,x=(3pi)/2