If sin⁡θ,cos⁡θ are the roots of equation 2x2-22x+1=0 then θ is equal to:

Given equation is,2x2-22x+1=0 Here, the roots are sinθ, cosθNow, for a standard quadratic equation ax2+bx+c=0,sum of roots=-ba and,Product of roots =ca, Thus in our case, sum of the roots,sin⁡θ+cos⁡θ=--222⇒sin⁡θ+cos⁡θ=2 …………. (1)Product of the roots,sin⁡θcos⁡θ=12⇒cos⁡θ=12sin⁡θ ………… (2)The new equation is produced by substituting the value from equation (2) into equation (1).sin⁡θ+cos⁡θ-cos⁡θ=2+12sin⁡θ ⇒2 2sin2⁡θ-22sin⁡θ+1=0 ⇒ (2sin⁡θ-1)2=0 (Since, a-b2=a2+b2-2ab)⇒sin θ =12 ⇒sin θ =sin⁡45oOn comparing we get,θ=45oThus, the value of θ=45 degrees. So, the correct option is 3.

If sin⁡θ,cos⁡θ are the roots of equation 2x2-22x+1=0 then θ is equal to:

Given equation is,2x2-22x+1=0 Here, the roots are sinθ, cosθNow, for a standard quadratic equation ax2+bx+c=0,sum of roots=-ba and,Product of roots =ca, Thus in our case, sum of the roots,sin⁡θ+cos⁡θ=--222⇒sin⁡θ+cos⁡θ=2 …………. (1)Product of the roots,sin⁡θcos⁡θ=12⇒cos⁡θ=12sin⁡θ ………… (2)The new equation is produced by substituting the value from equation (2) into equation (1).sin⁡θ+cos⁡θ-cos⁡θ=2+12sin⁡θ ⇒2 2sin2⁡θ-22sin⁡θ+1=0 ⇒ (2sin⁡θ-1)2=0 (Since, a-b2=a2+b2-2ab)⇒sin θ =12 ⇒sin θ =sin⁡45oOn comparing we get,θ=45oThus, the value of θ=45 degrees. So, the correct option is 3.