Question Number 191804 by Shlock last updated on 30/Apr/23
Commented by Shlock last updated on 30/Apr/23
x=?
Commented by AST last updated on 30/Apr/23
$${It}\:{follows}\:{from}\:{sin}^{\mathrm{2}} {x}+{cos}^{\mathrm{2}} {x}=\mathrm{1}. \\ $$
Answered by JDamian last updated on 30/Apr/23
$$\mathrm{2} \\ $$
Answered by mr W last updated on 30/Apr/23
$$\mathrm{cos}\:{x}\:{and}\:\mathrm{sin}\:\mathrm{2}\:{are}\:{side}\:{lengthes},\:{so} \\ $$$${they}\:{must}\:{be}\:{positive}\:{values}. \\ $$$$\mathrm{cos}\:{x}\:>\mathrm{0} \\ $$$$\mathrm{cos}^{\mathrm{2}} \:{x}+\mathrm{sin}^{\mathrm{2}} \:\mathrm{2}=\mathrm{1} \\ $$$$\mathrm{sin}\:\mathrm{2}=\mathrm{sin}\:\left(\pi−\mathrm{2}\right) \\ $$$$\mathrm{cos}\:{x}=\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:\mathrm{2}}=\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:\left(\pi−\mathrm{2}\right)}=\mathrm{cos}\:\left(\pi−\mathrm{2}\right) \\ $$$$\Rightarrow{x}=\left(\mathrm{2}{k}+\mathrm{1}\right)\pi\pm\mathrm{2}\:\checkmark \\ $$
Commented by mr W last updated on 01/May/23
