find this with steps fast please
I will mark in brainliest.if it is with steps
Answers
Step-by-step explanation:
Formula Used
\begin{gathered}\begin{gathered}\rm cos2\theta+1=2cos^2\theta\\\\\bullet \;\; \rm cos^2\theta=\dfrac{cos2\theta+1}{2}\\\\\bullet \;\; \rm cosA+cosB=\tiny{2cos(\dfrac{A+B}{2}).cos(\dfrac{A-B}{2})}\\\\\bullet \;\; \rm cos(\pi-\theta)=-sin\theta\end{gathered}\end{gathered}
cos2θ+1=2cos
2
θ
∙cos
2
θ=
2
cos2θ+1
∙cosA+cosB=2cos(
2
A+B
).cos(
2
A−B
)
∙cos(π−θ)=−sinθ
Solution
LHS
\begin{gathered}\begin{gathered}\rm \to cos^2x+cos^2(x+\dfrac{\pi}{3})+cos^2(x-\dfrac{\pi}{3})\\\\\end{gathered}\end{gathered}
→cos
2
x+cos
2
(x+
3
π
)+cos
2
(x−
3
π
)
\begin{gathered}\begin{gathered}\rm \to \dfrac{cos2x+1}{2}+\dfrac{\tiny{cos2(x+\dfrac{\pi}{3})+1}}{2}+\dfrac{\tiny{cos2(x-\dfrac{\pi}{3})+1}}{2}\\\\\to \rm \dfrac{1}{2}\tiny{( 3+cos2x+cos(2x+\dfrac{2\pi}{3})+cos(2x-\dfrac{2\pi}{3}))}\\\\\to \rm \dfrac{1}{2} \tiny{(3+cos2x+2cos(\dfrac{2x+\dfrac{2\pi}{3}+2x-\dfrac{2\pi}{3}}{2}).cos(\dfrac{2x+\dfrac{2\pi}{3}-2x+\dfrac{2\pi}{3}}{2}))}\\\\\end{gathered}\end{gathered}
→
2
cos2x+1
+
2
cos2(x+
3
π
)+1
+
2
cos2(x−
3
π
)+1
→
2
1
(3+cos2x+cos(2x+
3
2π
)+cos(2x−
3
2π
Answer:
₹60 is the answer
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