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# Solve for x: tan (x) + sec (x) = 2cos (x)?

Solve for x: tan (x) + sec (x) = 2cos (x)

### 6 Answers

- PuggyLv 71 decade agoFavorite Answer
tan(x) + sec(x) = 2cos(x)

Change everything to sines and cosines.

[sin(x)/cos(x)] + [1/cos(x)] = 2cos(x)

Get rid of all fractions by multiplying both sides by cos(x).

sin(x) + 1 = 2cos^2(x)

Change cos^2(x) into 1 - sin^2(x) as per the identity.

sin(x) + 1 = 2[1 - sin^2(x)]

sin(x) + 1 = 2 - 2sin^2(x)

Move everything to the left hand side.

2sin^2(x) + sin(x) + 1 - 2 = 0

2sin^2(x) + sin(x) - 1 = 0

Factor the same way you would factor 2y^2 + y - 1.

In fact, let me make it clear for you by using the substitution y = sin(x).

2y^2 + y - 1 = 0

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

Which means

2y - 1 = 0

2y = 1

y = 1/2, but back-substitute y = sin(x), to get

sin(x) = 1/2, so

x = { pi/6, 5pi/6 }

y + 1 = 0

y = -1

sin(x) = -1

x = 3pi/2

Therefore,

x = {pi/6, 5pi/6, 3pi/2 }

However, we should determine whether all of these solutions work, by plugging them into the original.

tan(x) + sec(x) = 2cos(x)

Test x = pi/6:

tan(pi/6) + sec(pi/6) = 2cos(pi/6)?

[1/sqrt(3)] + [2/sqrt(3)] = 2sqrt(3)/2 ?

[3/sqrt(3)] = sqrt(3) ?

3sqrt(3)/3 = sqrt(3) ?

sqrt(3) = sqrt(3) ? YES.

Keep this solution.

Test x = 5pi/6.

tan(5pi/6) + sec(5pi/6) = 2cos(5pi/6) ?

[-1/sqrt(3)] + [-2/sqrt(3)] = 2[-sqrt(3)/2] ?

-3/sqrt(3) = -sqrt(3) ?

-3sqrt(3)/3 = -sqrt(3) ?

-sqrt(3) = -sqrt(3) ? YES.

Test x = 3pi/2:

tan(x) + sec(x) = 2cos(x)

tan(3pi/2) + sec(3pi/2) = 2cos(3pi/2)

This one fails, because tan(3pi/2) is undefined.

Reject x = 3pi/2.

Making the only solutions

x = {pi/6, 5pi/6}

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- PhiloLv 71 decade ago
tan x + sec x = 2cos x

sin/cos + 1/cos = 2cos

sin + 1 = 2 cos²

sin + 1 = 2( 1 - sin²)

sin + 1 = 2 - 2sin²

2sin² + sin - 1 = 0

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

sin x = 1/2, x = π/6 or x = 5π/6

sin x = -1, x = 3π/2

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- 5 years ago
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RE:

Solve for x: tan (x) + sec (x) = 2cos (x)?

Solve for x: tan (x) + sec (x) = 2cos (x)

Source(s): solve tan sec 2cos x: https://tr.im/Ht6P3- Login to reply the answers

- NancieLv 44 years ago
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Pretty sure you can't divide by your du. But another way to approach it is expand it into 3/sec^2(x)+tan^2(x)/sec^2(x) sub trig id tan^2(x)=sec^2(x)-1 3/sec^2(x)+(sec^2(x)-1)/sec^2(x) expand second, 3/sec^2(x)+(sec^2(x)/sec^2(x))-1/sec^2(... 3/sec^2(x)-1/sec^2(x)+(sec^2(x)/sec^2(x... 2/sec^2(x)+1 sub trig id sec^2=1/cos^2(x) 2cos^2(x)+1 sub trig id cos^2(x)=(1/2)(1+cos2x) 2(1/2)(1+cos2x) +1 1+1+cos2x integrate 2+cos2x=2x+(1/2)sin2x Hope this helps

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