Question

poliedro medicina from hell

sabendo que 2 < x < 3 e que senx = 2raiz de 6sobre 7 entao cos de x eh igual a

a)5sobre 7
b)-5sobre 7
c)-3 sobre 7
D)-raiz de 37 sobre 7
e)raiz de 37 sobre 7

nao sei o gabarito resolucao pfvr

Answer 1

$\mathsf{(sen~x)^2+(cos~x)^2 = 1}\\\\\\ \mathsf{\left ( \dfrac{2\cdot \sqrt{6}}{7} \right )^2+(cos~x)^2 = 1}\\\\\\ \mathsf{\left (\dfrac{4\cdot 6}{49} \right) +(cos~x)^2 = 1}\\\\\\ \mathsf{\left (\dfrac{24}{49} \right) +(cos~x)^2 = 1}$

$\mathsf{(cos~x)^2 = 1-\left (\dfrac{24}{49} \right)}\\\\\\ \mathsf{(cos~x)^2 =\left (\dfrac{49-24}{49} \right)}\\\\\\ \mathsf{(cos~x)^2 =\left (\dfrac{25}{49} \right)}\\\\\\ \mathsf{cos~x =\sqrt{\dfrac{25}{49}}}$

$\boxed{\begin{array}{c} \mathsf{cos~x =\dfrac{5}{7}}\end{array}}$


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