Question

Solve the given equation in the interval [0,2 π]. Note: The answer must be written as a multiple of π. Give exact answers. Do not use decimal numbers. The answer must be an integer or a fraction. Note that π is already provided with the answer so you just have to find the appropriate multiple. Eg. if the answer is you should enter 1/2. If there is more than one answer write them separated by commas. 2(sin x)2-5 cos x + 1 = 0

Answer 1

Answer:

  x ∈ {π/3, 5π/3}

Step-by-step explanation:

  $2\sin^2{(x)}-5\cos{(x)}+1=0\\\\2(1-\cos^2{(x)})-5\cos{(x)}+1=0 \quad\text{use Pythagorean identity}\\\\-2\cos^2{(x)}-5\cos{(x)}+3=0 \quad\text{put in standard form}\\\\(\cos{(x)+3})(1-2\cos{(x)})=0 \quad\text{factor}\\\\\cos{(x)}=\dfrac{1}{2} \quad\text{use the zero product rule}\\\\x=\dfrac{1}{3}\pi\ \text{or}\ \dfrac{5}{3}\pi$