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Vesna [10]
3 years ago
14

Solve 2cos2x + cosx − 1 = 0 for x over the interval [0, 2pi ).

Mathematics
1 answer:
Galina-37 [17]3 years ago
4 0

2cos^2x+cosx-1 = 0

2cos^2x + 2cosx - cosx - 1 = 0

(2cos^2x + 2cosx) - (cosx + 1) = 0

2cosx(cosx + 1) - 1(cosx + 1) = 0

(cosx + 1)(2cosx - 1) = 0

cosx+1 = 0 or 2cosx-1 = 0

cosx+1 = 0 gives cosx = -1, x = PI

2cosx-1 = 0 gives cosx = 1/2, x = PI/3 or x = 5PI/3

answer: x = PI/3 or x = PI or x = 5PI/3
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Answer:

(a): <u>x</u><u> </u><u>is</u><u> </u><u>3</u><u> </u><u>and</u><u> </u><u>ky</u><u> </u><u>is</u><u> </u><u>-</u><u>1</u>

<u>(</u><u>b</u><u>)</u><u>:</u><u> </u><u>k</u><u> </u><u>is</u><u> </u><u>-</u><u>2</u>

Step-by-step explanation:

Let: 3x + ky = 8 be <em>e</em><em>q</em><em>u</em><em>a</em><em>t</em><em>i</em><em>o</em><em>n</em><em> </em><em>(</em><em>a</em><em>)</em>

x - 2 ky = 5 be <em>e</em><em>q</em><em>u</em><em>a</em><em>t</em><em>i</em><em>o</em><em>n</em><em> </em><em>(</em><em>b</em><em>)</em>

<em> </em>Then multiply <em>e</em><em>q</em><em>u</em><em>a</em><em>t</em><em>i</em><em>o</em><em>n</em><em> </em><em>(</em><em>a</em><em>)</em><em> </em>by 2:

→ 6x + 2ky = 16, let it be <em>e</em><em>q</em><em>u</em><em>a</em><em>t</em><em>i</em><em>o</em><em>n</em><em> </em><em>(</em><em>c</em><em>)</em>

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<em>{ \sf{(6  + 1)x + (2 - 2)ky = (16  +  5)}} \\ { \sf{7x = 21}} \\ { \sf{x = 3}}</em>

<em>T</em><em>h</em><em>e</em><em>n</em><em> </em><em>k</em><em>y</em><em> </em><em>:</em>

{ \sf{2ky = 3 - 5}} \\ { \sf{ky =  - 1}}

{ \bf{y =  \frac{1}{2} }} \\ { \sf{ky =  - 1}} \\ { \sf{k =  - 2}}

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