What is the general solution for tan(θ)=-(sqrt(7))/5 , pi/2 <θ<pi,cos(2θ) ?

The general solution for tan(θ)=-(sqrt(7))/5 , pi/2 <θ<pi,cos(2θ) is θ=-0.48669…+pi

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tan(θ)=-(sqrt(7))/5 , pi/2 <θ<pi,cos(2θ)

tan (θ) = - \frac{7}{5}, \frac{π}{2} < θ < π, cos (2 θ)

θ = - 0.48669 \dots + π

Solution steps

tan (θ) = - \frac{7}{5}, \frac{π}{2} < θ < π, cos (2 θ)

Apply trig inverse properties

θ = arctan (- \frac{7}{5}) + πn

Solutions for the range

\frac{π}{2} < θ < π

θ = - arctan (\frac{7}{5}) + π

Show solutions in decimal form

θ = - 0.48669 \dots + π

cos (θ) = - (\frac{1}{2})

cos (θ) = 2 cos (θ) - 1

2 cos^{2} (2 β) + cos (2 β) - 1 = 0, - 36 0^{\circ} \leq β \leq 36 0^{\circ}

cos (2 x) = \frac{3}{7}

0 = 1 - 3 cos (θ)

What is the general solution for tan(θ)=-(sqrt(7))/5 , pi/2 <θ<pi,cos(2θ) ?
The general solution for tan(θ)=-(sqrt(7))/5 , pi/2 <θ<pi,cos(2θ) is θ=-0.48669…+pi