Question

Cos theta +cos3theta +2 cos 2theta=0

Answer 1

Given,

$\longrightarrow\sf{\cos\theta+\cos(3\theta)+2\cos(2\theta)=0}$

Or,

$\longrightarrow\sf{\left[\cos\theta+\cos(3\theta)\right]+2\cos(2\theta)=0}$

We know the sum - to - product identity,

$\longrightarrow\sf{\cos A+\cos B=2\cos\left(\dfrac{A+B}{2}\right)\cos\left(\dfrac{A-B}{2}\right)}$

Then,

$\longrightarrow\sf{2\cos\left(\dfrac{\theta+3\theta}{2}\right)\cos\left(\dfrac{\theta-3\theta}{2}\right)+2\cos(2\theta)=0}$

$\longrightarrow\sf{2\cos\left(2\theta\right)\cos\left(-\theta\right)+2\cos(2\theta)=0}$

Since $\sf{\cos(-\theta)=\cos\theta,}$

$\longrightarrow\sf{2\cos\left(2\theta\right)\cos\theta+2\cos(2\theta)=0}$

$\longrightarrow\sf{2\cos\left(2\theta\right)\big[\cos\theta+1\big]=0}$

This implies,

$\longrightarrow\sf{\cos(2\theta)=0\quad\quad OR\quad\quad \cos\theta=-1}$

$\Longrightarrow\sf{2\theta=\dfrac{(2n+1)\pi}{2}\quad\quad OR\quad\quad \theta=(2n+1)\pi}$

$\longrightarrow\sf{\underline{\underline{\theta=\dfrac{(2n+1)\pi}{4}}}\quad\quad OR\quad\quad\underline{\underline{\theta=(2n+1)\pi}}}$

for every $\sf{n\in\mathbb{Z}.}$

Answer 2

Given : Cosθ + Cos3θ  + 2Cos2θ = 0

To find : Value of θ (  theta)

Solution:

Cosθ + Cos3θ  + 2Cos2θ = 0

using  Cos3θ = 4Cos³θ - 3Cosθ    & Cos2θ = 2Cos²θ - 1

=> Cosθ +  4Cos³θ - 3Cosθ + 2( 2Cos²θ - 1) = 0

=> Cosθ +  4Cos³θ - 3Cosθ + 4Cos²θ - 2 =0

=>  4Cos³θ  + 4Cos²θ - 2Cosθ  - 2 = 0

=>  2Cos³θ  + 2Cos²θ -  Cosθ  - 1 = 0

=> 2Cos²θ(Cosθ + 1) - 1(Cosθ + 1) = 0

=> (  2Cos²θ- 1 ) (Cosθ + 1)  = 0

=> Cos²θ = 1/2   or Cosθ = -1

=> Cosθ = ± 1/√2  , - 1

θ = (2n + 1)π/4  , (2n + 1)π

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