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

The general solution for 2sin^2(x)+5cos(x)-3=0 is x=1.34977…+2pin,x=2pi-1.34977…+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2sin^2(x)+5cos(x)-3=0

2 sin^{2} (x) + 5 cos (x) - 3 = 0

x = 1.34977 \dots + 2 πn, x = 2 π - 1.34977 \dots + 2 πn

Solution steps

2 sin^{2} (x) + 5 cos (x) - 3 = 0

Rewrite using trig identities

- 1 - 2 cos^{2} (x) + 5 cos (x) = 0

Solve by substitution

cos (x) = \frac{5 - 17}{4}, cos (x) = \frac{5 + 17}{4}

cos (x) = \frac{5 - 17}{4} : x = arccos (\frac{5 - 17}{4}) + 2 πn, x = 2 π - arccos (\frac{5 - 17}{4}) + 2 πn

cos (x) = \frac{5 + 17}{4} :

No Solution

Combine all the solutions

x = arccos (\frac{5 - 17}{4}) + 2 πn, x = 2 π - arccos (\frac{5 - 17}{4}) + 2 πn

Show solutions in decimal form

x = 1.34977 \dots + 2 πn, x = 2 π - 1.34977 \dots + 2 πn

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

sin (θ) = - \frac{27}{190}

4 sin (θ) = 1 + sin (θ)

tan (θ) - 3 = 0, 0 \leq θ \leq 2 π

\frac{1}{32} \cdot 120 0^{2} \cdot sin (2 θ) = 3000

What is the general solution for 2sin^2(x)+5cos(x)-3=0 ?
The general solution for 2sin^2(x)+5cos(x)-3=0 is x=1.34977…+2pin,x=2pi-1.34977…+2pin