Question

Solve the given equation. (Enter your answers as a comma-separated list. Let k be any integer. Round terms to three decimal places where appropriate. If there is no solution, enter NO SOLUTION.)

cos²(θ) − cos(θ) − 12 = 0

Answer 1

For the polynomial ax2 + bx + c =0 ,

The value of x is given by

x = (-b √( b2 - 4ac) ) / 2a

Some trigonometry formula

cos 2Φ = 2 cos2 Φ - 1

And the value of cosine is always lie between -1 and 1 .

That is, -1 cos Φ 1. For every Φ € R

Given cos 2 - cos -12 = 0

2 cos2 - 1 - cos -12 = 0

2 cos2 - cos -13 = 0

cos = {-(-1) √( (-1)2 - 4 (2)(-13) ) }/ 2(2)

cos = { 1 √ ( 1 + 104) } / 4 = { 1 √105 } /4 = { 1 10.247 } /4

cos = (1 + 10.247)/4 or (1-10.247)/4

cos = 2.812 or -2.312

But cos always lies between -1 and 1. And since both values are not between -1 and 1.

We come to the conclusion that cos doesn't exist for the given equation.

Thus, no solution exists for the given equation..

Problem has been solved as per your wish....