What is the general solution for 4sin^2(x)+4cos(x)-5=0,0<= x<= 2pi ?

The general solution for 4sin^2(x)+4cos(x)-5=0,0<= x<= 2pi is x= pi/3 ,x=(5pi)/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4sin^2(x)+4cos(x)-5=0,0<= x<= 2pi

4 sin^{2} (x) + 4 cos (x) - 5 = 0, 0 \leq x \leq 2 π

x = \frac{π}{3}, x = \frac{5 π}{3}

Solution steps

4 sin^{2} (x) + 4 cos (x) - 5 = 0, 0 \leq x \leq 2 π

Rewrite using trig identities

- 1 + 4 cos (x) - 4 cos^{2} (x) = 0

Solve by substitution

cos (x) = \frac{1}{2}

cos (x) = \frac{1}{2}, 0 \leq x \leq 2 π : x = \frac{π}{3}, x = \frac{5 π}{3}

Combine all the solutions

x = \frac{π}{3}, x = \frac{5 π}{3}

2 sin (2 θ) = 1

cos (θ) = \frac{1}{2}, 0 \leq θ \leq 4 π

sin (x) a = \frac{π}{2}

tan^{2} (3 θ) = 1

tan^{2} (2 θ) - 3 = 0

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