Let sin(2x) = cos(x), where 0° ≤ x < 180°. what are the possible values for x? 30° only 90° only 30° or 150° 30°, 90°, or 150°
Let Sin(2X) = Cos(X), Where 0° ≤ X < 180°. What Are The Possible Values For X? 30° Only 90° Only 30° Or 150° 30°, 90°, Or 150°
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Let Sin(2X) = Cos(X), Where 0° ≤ X < 180°. What Are The Possible Values For X? 30° Only 90° Only 30° Or 150° 30°, 90°, Or 150°. What are the possible values for x? Click here 👆 to get an answer to your question ️ let sin (2x)=cos (x) , where 0°≤ x<180°.
Ex 3.4, 7 Find general solution of sin 2x + cos x = 0 Ex 3.4 from www.teachoo.com
The solutions to the equation sin(2x) = cos(x) where 0∘ ≤ x <180∘ are x = 30∘, x = 90∘, and x = 150∘. The solution to the equation sin (2x) = cos (x), within the range of 0° to 180°, gives us potential values of 30° and 150° for x, corresponding to the answer option c. What are the possible values for x?
Step 1/3First, We Can Use The Identity Sin (2X) = 2Sin (X)Cos (X) To Rewrite The Equation As:
I changed it to cos(x) = sin(x) cos(x) − cos(x) sin(x) cos (x) = sin (x) cos (x) − cos (x) sin (x). 30° only 90° only 30° or 150° 30 2sin (x)cos (x) = cos (x)step 2/3next, we can divide both sides by cos (x) (since cos (x) cannot be.
The Solutions To The Equation Sin(2X) = Cos(X) Where 0∘ ≤ X <180∘ Are X = 30∘, X = 90∘, And X = 150∘.
Click here 👆 to get an answer to your question ️ let sin (2x)=cos (x) , where 0°≤ x<180°. Using trigonometric identities and properties, these solutions are derived. Find the exact values of x x so that 0 ≤ x <2π 0 ≤ x <2 π and cos(x) = sin(2x) cos (x) = sin (2 x).
The Solution To The Equation Sin (2X) = Cos (X), Within The Range Of 0° To 180°, Gives Us Potential Values Of 30° And 150° For X, Corresponding To The Answer Option C.
What are the possible values for x? Since the sine function is periodic, we.