What is the general solution for 8cos^2(θ)-3cos(θ)=0,0<= θ<= 180 ?

The general solution for 8cos^2(θ)-3cos(θ)=0,0<= θ<= 180 is θ=1.18639…,θ=90

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

8cos^2(θ)-3cos(θ)=0,0<= θ<= 180

8 cos^{2} (θ) - 3 cos (θ) = 0, 0^{\circ} \leq θ \leq 18 0^{\circ}

θ = 1.18639 \dots, θ = 9 0^{\circ}

Solution steps

8 cos^{2} (θ) - 3 cos (θ) = 0, 0^{\circ} \leq θ \leq 18 0^{\circ}

Solve by substitution

cos (θ) = \frac{3}{8}, cos (θ) = 0

cos (θ) = \frac{3}{8}, 0 \leq θ \leq 18 0^{\circ} : θ = arccos (\frac{3}{8})

cos (θ) = 0, 0 \leq θ \leq 18 0^{\circ} : θ = 9 0^{\circ}

Combine all the solutions

θ = arccos (\frac{3}{8}), θ = 9 0^{\circ}

Show solutions in decimal form

θ = 1.18639 \dots, θ = 9 0^{\circ}

8 sin (x) - 7 = 0

1 - sin (θ) = 1 + cos (θ)

sec (t) = 2

tan (x + 2) = cot (4 x + 28)

12 \cdot sin (3 x) = 0

What is the general solution for 8cos^2(θ)-3cos(θ)=0,0<= θ<= 180 ?
The general solution for 8cos^2(θ)-3cos(θ)=0,0<= θ<= 180 is θ=1.18639…,θ=90