trig
posted by Sierra .
Solve the equation on the interval [0,360) cos^2t+2cos(t)+1=0

cos^2t+2cos(t)+1=0
(cost + 1)^2 = 0
cost + 1 = 0
cost = 1
t = 270°
Respond to this Question
Similar Questions

math (final trig problem)
Solve for x on the interval [0,π/2): cos^3(2x) + 3cos^2(2x) + 3cos(2x) = 4 I havent done trig for a while so what exactly does that mean in solving for x on that interval and how would i go about doing that? 
Math  Solving for Trig Equations
Solve the following equation for 0 less than and/or equal to "x" less than and/or equal to 360  cos^2x  1 = sin^2x  Attempt: cos^2x  1  sin^2x = 0 cos^2x  1  (1  cos^2x) = 0 cos^2x  1  1 + cos^2x = 0 2cos^2x  2 = 0 (2cos^2x/2)= … 
Math  Solving Trig Equations
What am I doing wrong? Equation: sin2x = 2cos2x Answers: 90 and 270 .... My Work: 2sin(x)cos(x) = 2cos(2x) sin(x) cos(x) = cos(2x) sin(x) cos(x) = 2cos^2(x)  1 cos(x) (+/)\sqrt{1  cos^2(x)} = 2cos^2(x)  1 cos^2(x)(1  cos^2(x)) 
trig
find in degrees the value of Ө in the interval O¡ÜӨ¡Ü 360 for which 2cos^2ӨcosӨ1=sin^2Ө give your answers to 1 decimal place where appropriate 
algebra 2 and trig (math)
Find the value of x in the interval 0 is less than or equal to x and 360 is greater than or equal to x which satisfies the equation cos x  2 cos x sin x. 
MathsSs triG
Consider sin(x360)sin(90x)tan(x)/cos(90+x) 1.A.SIMPLIFY sin(x360)sin(90x)tan(x)/cos(90+x) to a single trigonometric ratio B.hence or otherwise without using a calculator,solve for X if 0<X<360. sin(x360)sin(90x)tan(x)/cos(90+x) … 
MathS triG
1 .If tanA=1/3 and tanB=1/7 (both A and B are acute),calculate 50sin(2A +B) 2.1Prove that sin2A+2cosA2cos^3A/1+sinA =sin2A 2.2 for which values of A in the interval[360;360] is the identity in 2.1 undefined? 
maths Pls help trig
1 .If tanA=1/3 and tanB=1/7 (both A and B are acute),calculate 50sin(2A +B) 2.1Prove that sin2A+2cosA2cos^3A/1+sinA =sin2A 2.2 for which values of A in the interval[360;360] is the identity in 2.1 undefined? 
trignometry
Give a list of solutions, in the interval [0, 360°), in degrees, of the equation 2cos^24θ + 5 cos 4θ = 3. 
Trig
Find all solutions of the equation in the interval [0,2pi) 2 cos^2 xcos x = 0 2cos^2 + cosx + 0 (x+1/2) (x+0/2) (2x+1) (x+0) 1/2,0 2Pi/3, 4pi/3, pi/2, 3pi/2 my teacher circled pi/2 and 3pi/2 What did I do wrong?