$\frac{ {tan}^{2}x }{2} - 2 {cos}^{2}x = 1 \\ \\ \frac{ {tan}^{2}x - 4{cos}^{2}x }{2} = 1 \\ \\ {tan}^{2}x - 4{cos}^{2}x = 2 \\ \\ \frac{{sin}^{2}x}{{cos}^{2}x} - 4{cos}^{2}x = 2 \\ \\ \frac{{sin}^{2}x - 4{cos}^{4}x}{{cos}^{2}x} = 2 \\ \\ {sin}^{2}x - 4{cos}^{4}x = 2{cos}^{2}x \\ \\ 4{cos}^{4}x + 2{cos}^{2}x - {sin}^{2}x = 0 \\ \\ 4{cos}^{4}x + 2{cos}^{2}x + {cos}^{2}x - 1= 0 \\ \\ 4{cos}^{4}x + 3{cos}^{2}x - 1= 0 \\ \\ 4{cos}^{4}x + 4{cos}^{2}x - {cos}^{2}x - 1= 0 \\ \\4{cos}^{2}x({cos}^{2}x + 1) - 1({cos}^{2}x + 1) = 0 \\ \\ ({cos}^{2}x + 1)(4{cos}^{2}x - 1) = 0 \\ \\ ({cos}^{2}x + 1) = 0 \: or \: (4{cos}^{2}x - 1) = 0 \\ \\ {cos}^{2}x = - 1 \: or \: 4{cos}^{2}x = 1 \\ \\ {cos}x = \sqrt{ - 1} \: which \: is \: not \: possible \\ \therefore \: {cos}^{2}x = \frac{1}{4} \\ \\ \therefore \: {cos}x = \pm\frac{1}{2} \\ \\ \therefore \: {cos}x = \frac{1}{2} \: or \: {cos}x = - \frac{1}{2} \\ \\ \therefore \: {cos}x = {cos}60 \degree \: or \: {cos}x = {cos}120 \degree \\ \\ \therefore \:x = 60 \degree \: \: or \: \: x = 120 \degree$

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3 years ago