Mathematics, 04.02.2020 09:55, Quakersoatmeal9
Verify that the given value is a solution for the trigonometric equation by showing that it makes the equation true:
1. 2 cos^2 x-cos x=0; 2π/3
2. 2 cos^2 x-cos x=0; 0
Answers: 2
![\bf \begin{array}{llll} 2cos^2(x)-cos(x)&=&0\\\\ 2cos^2\left( \frac{2\pi }{3} \right)-cos\left( \frac{2\pi }{3} \right)\\\\ 2\left[ cos\left( \frac{2\pi }{3} \right) \right]^2 -cos\left( \frac{2\pi }{3} \right)\\\\ 2\left[ -\frac{1}{2} \right]^2-\left( -\frac{1}{2} \right)\\\\ 2\left[ \cfrac{1}{4} \right]+\cfrac{1}{2}\\\\ \cfrac{1}{2}+\cfrac{1}{2}\\\\ 1&\ne & 0 \end{array} \\\\[-0.35em] ~\dotfill](/tpl/images/0499/9950/39a54.png)
![\bf \begin{array}{llll} 2cos^2(x)-cos(x)&=&0\\\\ 2cos^2(0)-cos(0)\\\\ 2[cos(0)]^2-cos(0)\\\\ 2[1]^2-1\\\\ 2-1\\\\ 1&\ne & 0 \end{array}](/tpl/images/0499/9950/2aafc.png)
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