Question

3 sin2x + 2cos2x = 3​

Answer 1

Пользуемся формулой введения вспомогательного угла:

$a sin alpha + b cos alpha = sqrt{ {a}^{2} + {b}^{2} } sin( alpha + gamma ), : gamma = arctg frac{b}{a} = arcsin frac{b}{ sqrt{ {a}^{2} + {b}^{2} }}= arccos frac{a}{ sqrt{ {a}^{2} + {b}^{2} }}$

$3 sin 2x + 2 cos 2x = 3 \ \ sqrt{ {3}^{2} + {2}^{2} } sin(2x + gamma ) = 3, : gamma = arcsin frac{2}{ sqrt{13} } \ \ sqrt{13} sin(2x + arcsin frac{2}{ sqrt{13} } ) = 3 \ \ sin(2x + arcsin frac{2}{ sqrt{13} } ) = frac{3}{ sqrt{13} } \ \ 2x + arcsin frac{2}{ sqrt{13} } = ( - 1) ^{n} arcsin frac{3}{ sqrt{13} } + pi n \ \ 2x = ( - 1) ^{n} arcsin frac{3}{ sqrt{13} } - arcsin frac{2}{ sqrt{13} } + pi n \ \ x = ( - 1) ^{n} times frac{1}{2} arcsin frac{3}{ sqrt{13} } - frac{1}{2} arcsin frac{2}{ sqrt{13} } + frac{1}{2} pi n \ \ OTBET: ( - 1) ^{n} times frac{1}{2} arcsin frac{3}{ sqrt{13} } - frac{1}{2} arcsin frac{2}{ sqrt{13} } + frac{1}{2} pi n, : n in Z$