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Sin(x-20)=1/root2

This question is confusing me.

Sagot :

Crisforpid
=> x - π/9 = π/4 + 2kπ, where k is an integer or  x - π/9 =  3π/4  + 2kπ, where k is an integer => x = 13π/36 +  2kπ, where k is an integer or x = 31π/36 + 2kπ, where k is an integer;
GrandNecroid

Answer:

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Step-by-step explanation:

Note that 1/√(2) is equal to √(2)/2:

1/√(2) = √(2)/2

So we're given that sin(x-20) = √(2)/2

The only place on the unit circle where sin() is equal to √(2)/2 is at π/4 and 3π/4.

That means the value inside the sine must be equal to π/4 and 3π/4:

since sin(π/4) = √(2)/2  and  sin(π/4) = √(2)/2

then,

sin(π/4) = sin(x-20)   and  sin(π/4) = sin(x-20)

You set the inside of the parentheses equal to each other and solve for x.

You can also add a +2πk at the end since all sinusoidal graph repeats every 2π.

View image GrandNecro

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