Question

(2cosx-√2)(√sinx+1)=0

Answer 1

$(2cosx-\sqrt2)(\sqrt{sinx+1})=0\\\\a)\; \; cox=\frac{\sqrt2}{2}\\\\x=\pm \frac{\pi }{4} +2\pi n\; ,\; n\in Z\\\\b)\; \; \sqrt{sinx+1}=0\; \; ,\; \; sinx+1=0\; \; ,\; \; sinx=-1\\\\x=- \frac{\pi }{2}+2\pi k,\; k\in Z\\\\Otvet:\; \; x=\pm \frac{\pi }{4}+2\pi n\; ,\; \; x=-\frac{\pi }{2}+2\pi k\; ,\; \; n,k\in Z\; .$

Answer 2

To solve this equation, we need to find the values of x that make the left-hand side equal to zero, since the product of two factors is zero only when at least one of them is zero.

So, we have two factors: (2cosx-√2) and (√sinx+1). To find the values of x that make each factor zero, we set each factor equal to zero and solve for x:

2cosx-√2 = 0 cosx = √2/2 x = π/4 + kπ or x = 7π/4 + kπ where k is any integer

√sinx+1 = 0 sinx = -1 x = 3π/2 + kπ where k is any integer

Therefore, the solutions to the given equation are: x = π/4 + kπ, 7π/4 + kπ or x = 3π/2 + kπ where k is any integer.