phương trình lượng giác không mẫu mực

K

khongphaibang

Câu 1 : ${\cos ^{13}}x + {\sin ^{14}}x = 1$

Ta có : Do $\sin x,\cos x \in [ - 1;1]$ nên ta có :

${\cos ^{13}}x \le {\cos ^2}x,{\sin ^{14}}x \le {\sin ^2}x$

Vậy :${\cos ^{13}}x + {\sin ^{14}}x \le {\cos ^2}x + {\sin ^2}x = 1$

Do đó : dấu bằng xảy ra khi và chỉ khi

$\left\{ \begin{array}{l}
{\sin ^{14}}x = {\sin ^2}x\\
{\cos ^{13}}x = {\cos ^2}x
\end{array} \right.$
 
C

conga222222

$\eqalign{
& 1, \cr
& co\;1 - co{s^{13}}x - {\sin ^{14}}x = {\cos ^2}x - {\cos ^{13}}x + {\sin ^2}x - {\sin ^{14}}x \cr
& = {\cos ^2}x\left( {1 - {{\cos }^{11}}x} \right) + {\sin ^2}x\left( {1 - {{\sin }^{12}}x} \right) \geqslant 0\;\left( {do\;sinx \leqslant 1\;;cosx \leqslant 1} \right) \cr
& dau = \leftrightarrow .... \cr
& 2) \cr
& {2^{\left| {\sin x} \right|}} \geqslant 1 \geqslant \cos x\;\left( {do\;\left| {\sin x} \right| \geqslant 0} \right) \cr
& \left| {\sin x} \right| - {\sin ^2}x = \left| {\sin x} \right|\left( {1 - \left| {\sin x} \right|} \right) \geqslant 0 \cr
& \to {2^{\left| {\sin x} \right|}} + \left| {\sin x} \right| \geqslant {\sin ^2}x + \cos x \cr
& dau = \leftrightarrow .... \cr
& 3) \cr
& co\;sinx \geqslant 0\;\forall x \in \left[ {0;\frac{\pi }{2}} \right] \cr
& \to {2^{2\sin x}} \geqslant 1 \geqslant \cos x \cr
& dau = \leftrightarrow .... \cr} $
 
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