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Tasya [4]
2 years ago
10

Sin^2x + sin2x +2cox^2x = 2

Mathematics
1 answer:
Effectus [21]2 years ago
5 0

Answer:

Step-by-step explanation:

sin^2 \ x + sin ^2 \ x + 2cos^ 2\ x = 2\\\\2sin^2 \ x  + 2 cos^2 \ x = 2\\\\2 ( sin^2 \ x + cos^2 \ x ) = 2 \\\\2 \times 1 = 2 \\\\2 = 2 \\\\LHS = RHS \\\\Hence \ proved.

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lubasha [3.4K]

The zeros of given function y=x^{2}+8 x+15 is – 5 and – 3

<u>Solution:</u>

\text { Given, equation is } y=x^{2}+8 x+15

We have to find the zeros of the function by rewriting the function in intercept form.

By using intercept form, we can put value of y as  to obtain zeros of function

We know that, intercept form of above equation is x^{2}+8 x+15=0

\text { Splitting } 8 x \text { as }(5+3) x \text { and } 15 \text { as } 5 \times 3

\begin{array}{l}{\rightarrow x^{2}+(5+3) x+5 \times 3=0} \\\\ {\rightarrow x^{2}+5 x+3 x+5 \times 3=0}\end{array}

Taking “x” as common from first two terms and “3” as common from last two terms

x (x + 5) + 3(x + 5) = 0

(x + 5)(x + 3) = 0

Equating to 0 we get,

x + 5 = 0 or x + 3 = 0

x = - 5 or – 3

Hence, the zeroes of the given function are – 5 and – 3

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