Given: 2sin4(α)+2cos4(α)−1=0, where 0<α<π∕2 Formula used: sin4(α)+cos4(α)=(sin2(α)+cos2(α))2−2sin2(α)cos2(α) We know that sin2(α)+cos2(α)=1. Calculation: The given equation is: 2sin4(α)+2cos4(α)−1=0 ⇒2(sin4(α)+cos4(α))=1 ⇒sin4(α)+cos4(α)=1∕2 Now, using the identity: sin4(α)+cos4(α)=(sin2(α)+cos2(α))2−2sin2(α)cos2(α) ⇒(1)2−2sin2(α)cos2(α)=1∕2 ⇒1−2sin2(α)cos2(α)=1∕2 ⇒2sin2(α)cos2(α)=1∕2 ⇒sin2(α)cos2(α)=1∕4 Now, take the square root of both sides: ⇒sin(α)cos(α)=1∕2 Using the identity sin(2α)=2sin(α)cos(α) : ⇒sin(2α)=2×1∕2=1 sin(2α)=1 implies that: 2α=90∘ Therefore: α=45∘ sin(2α)=sin(90∘)=1 cos(2α)=cos(90∘)=0 Thus: sin(2α)+cos(2α)=1+0=1. sin(2α)+cos(2α)=1
