Ответ:
\begin{gathered}\sin^2(x)+3\sin(x)\cos(x)-4\cos^2(x)=0\Big/\div\cos^2(x)\\\tan^2(x)+3\tan(x)-4=0\\\tan(x)=u\\u^2+3u-4=0\\u_{1,2}=1;-4\\\begin{cases}\tan(x)=1\\\tan(x)=-4\end{cases}\Rightarrow\begin{cases}x=\frac{\pi}{4}+\pi k\\x=-\tan^{-1}(4)+\pi k\end{cases}, k\in \mathbb Z\end{gathered}
sin
2
(x)+3sin(x)cos(x)−4cos
(x)=0/÷cos
(x)
tan
(x)+3tan(x)−4=0
tan(x)=u
u
+3u−4=0
1,2
=1;−4
{
tan(x)=1
tan(x)=−4
⇒{
x=
4
π
+πk
x=−tan
−1
(4)+πk
,k∈Z
P.s тангенс в минус первой степени - это арктангенс.
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Answers & Comments
Ответ:
\begin{gathered}\sin^2(x)+3\sin(x)\cos(x)-4\cos^2(x)=0\Big/\div\cos^2(x)\\\tan^2(x)+3\tan(x)-4=0\\\tan(x)=u\\u^2+3u-4=0\\u_{1,2}=1;-4\\\begin{cases}\tan(x)=1\\\tan(x)=-4\end{cases}\Rightarrow\begin{cases}x=\frac{\pi}{4}+\pi k\\x=-\tan^{-1}(4)+\pi k\end{cases}, k\in \mathbb Z\end{gathered}
sin
2
(x)+3sin(x)cos(x)−4cos
2
(x)=0/÷cos
2
(x)
tan
2
(x)+3tan(x)−4=0
tan(x)=u
u
2
+3u−4=0
u
1,2
=1;−4
{
tan(x)=1
tan(x)=−4
⇒{
x=
4
π
+πk
x=−tan
−1
(4)+πk
,k∈Z
P.s тангенс в минус первой степени - это арктангенс.