due apr 4 by 11:59pm points 15 submitting available until may 2 at 11:59pm hw 8.4: solving hardest…
due apr 4 by 11:59pm points 15 submitting available until may 2 at 11:59pm hw 8.4: solving hardest trigonometric score: 3.1/15 answered: 4/15 question 4 score on last try: 0.1 of 1 pts. see details for more. at least one scored part is incorrect. jump to first change > next question get a similar question solve 2cos²(x) - 7cos(x) + 5 = 0 for all solutions. x = 0 where k is any integer give your answer as an exact value. question help: video
Answer
Explanation:
Step1: Let $t = \cos(x)$
The equation becomes $2t^{2}-7t + 5=0$.
Step2: Factor the quadratic equation
We have $(2t - 5)(t - 1)=0$.
Step3: Solve for $t$
Set each factor equal to zero: For $2t-5 = 0$, $t=\frac{5}{2}$; for $t - 1=0$, $t = 1$.
Step4: Recall the range of cosine
Since $- 1\leqslant\cos(x)\leqslant1$, $t=\frac{5}{2}$ is not a valid solution as $\frac{5}{2}>1$.
Step5: Solve for $x$ when $\cos(x)=1$
We know that $\cos(x)=1$ when $x = 2k\pi$, where $k\in\mathbb{Z}$.
Answer:
$x = 2k\pi$, where $k$ is any integer.