Please help with this problem.

Mathematics

1 answer:

larisa86 [58]3 years ago

6 0

Answer:

60°, 120°

Step-by-step explanation:

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

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What is one third the difference between 3/4 and 2/3?

Ivan

Answer:

One third the difference between 3/4 and 2/3 is $(\frac{1}{36} )$

Step-by-step explanation:

One third the difference of A and B = $\frac{1}{3} \times (A-B)$

Here, as given in the question:

$A = \frac{3}{4} , B = \frac{2}{3} \\\implies A - B = \frac{3}{4} - \frac{2}{3}\\=(\frac{3 \times 3 -2 \times 4}{12}) = (\frac{9-8}{12} ) = (\frac{1}{12})\\ \implies A - B = (\frac{1}{12})$