Solution (By Examveda Team)
$$\eqalign{
& {\text{7co}}{{\text{s}}^2}\theta + 3{\sin ^2}\theta = 4 \cr
& 7{\text{co}}{{\text{s}}^2}\theta + 3\left( {1 - {\text{co}}{{\text{s}}^2}\theta } \right) - 4 = 0 \cr
& 7{\text{co}}{{\text{s}}^2}\theta - 3{\text{co}}{{\text{s}}^2}\theta + 3 - 4 = 0 \cr
& 4{\text{co}}{{\text{s}}^2}\theta = 1 \cr
& {\text{co}}{{\text{s}}^2}\theta = \frac{1}{4} \cr
& {\text{cos}}\theta = \frac{1}{2} \cr
& \cos \theta = {\text{cos}}{60^ \circ } \cr
& \theta = {60^ \circ } \cr} $$
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