Równania...

[siua17]

Jak to obliczyć? I co robię nie tak? Proszę o wyraźne obliczenia.

Działanie: \(\displaystyle{ 2cos^{2}x+4sin^{2}x=3}\)
Moje próby rozwiązania:
\(\displaystyle{ 2cos^{2}x+4(1-cos^{2}x)=3}\)
\(\displaystyle{ 2cos^{2}x+4-4cos^{2}x=3}\)
\(\displaystyle{ 2cos^{2}x-4cos^{2}x=3-4}\)
\(\displaystyle{ -2cos^{2}x=-1 /:(-2)}\)
\(\displaystyle{ cos^{2}x=\frac{1}{2}}\)
\(\displaystyle{ cosx=\sqrt{\frac{1}{2}}*\frac{\sqrt{2}}{\sqrt{2}}}\)
\(\displaystyle{ cosx=\frac{\sqrt{2}}{2}}\)
\(\displaystyle{ x_{0}=\frac{\pi}{4}}\)
\(\displaystyle{ x_{1}=\frac{\pi}{4}+2k\pi; k \in C}\)
\(\displaystyle{ x_{2}=-\frac{\pi}{4}+2k\pi; k \in C}\)

[Mikolaj9]

\(\displaystyle{ cos ^{2}x= \frac{1}{2}}\)
\(\displaystyle{ cosx= \frac{ \sqrt{2} }{2} \vee cosx= -\frac{ \sqrt{2} }{2}}\)

I.
\(\displaystyle{ x_{1}= \frac{ \pi}{4}+2k \pi; k \in C}\)
\(\displaystyle{ x_{2}=- \frac{ \pi}{4}+2k \pi; k \in C}\)

II.
\(\displaystyle{ cos(\pi-x)= \frac{ \sqrt{2} }{2}}\)
\(\displaystyle{ x _{3}=...}\)
\(\displaystyle{ x _{4}=...}\)